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NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry | EduGrown

In This Post we are  providing Chapter 3 Coordinate Geometry NCERT Solutions for Class 9 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Coordinate Geometry Class 9 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 9 Maths Coordinate Geometry NCERT Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry

https://youtu.be/qezAU35eZm0

Exercise 3.1

1. How will you describe the position of a table lamp on your study table to another person?

Answer

To describe the position of a table lamp on the study table, we have two take two lines, a perpendicular and horizontal. Considering the table as a plane and taking perpendicular line as Y axis and horizontal as X axis. Take one corner of table as origin where both X and Y axes intersect each other. Now, the length of table is Y axis and breadth is X axis. From The origin, join the line to the lamp and mark a point. Calculate the distance of this point from both X and Y axes and then write it in terms of coordinates.
Let the distance of point from X axis is x and from Y axis is y then the the position of the table lamp in terms of coordinates is (x,y).


2. (Street Plan) : A city has two main roads which cross each other at the centre of the city. These two roads are along the North-South direction and East-West direction.
All the other streets of the city run parallel to these roads and are 200 m apart. There are 5 streets in each direction. Using 1cm = 200 m, draw a model of the city on your notebook. Represent the roads/streets by single lines. There are many cross- streets in your model. A particular cross-street is made by two streets, one running in the North – South direction and another in the East – West direction. Each cross street is referred to in the following manner : If the 2nd street running in the North – South direction and 5 th in the East – West direction meet at some crossing, then we will call this cross-street (2, 5). Using this convention, find:
(i) how many cross – streets can be referred to as (4, 3).
(ii) how many cross – streets can be referred to as (3, 4)

Answer

 
(i) Only one street can be referred to as (4, 3) as we see from the figure.
(ii) Only one street can be referred to as (3, 4) as we see from the figure.

https://youtu.be/usRYLKnTT_E

Exercise 3.2
 
1. Write the answer of each of the following questions:

(i) What is the name of horizontal and the vertical lines drawn to determine the position of any point in the Cartesian plane?

(ii) What is the name of each part of the plane formed by these two lines?
(iii) Write the name of the point where these two lines intersect.
 
Answer
 
(i) The name of horizontal lines and vertical lines drawn to determine the position of any point in the Cartesian plane is x-axis and y-axis respectively.
(ii) The name of each part of the plane formed by these two lines x-axis and y-axis is quadrants.
(iii) The point where these two lines intersect is called origin.
 
 
 
2. See Fig.3.14, and write the following:

(i) The coordinates of B.
(ii) The coordinates of C.
(iii) The point identified by the coordinates (-3, -5).
(iv) The point identified by the coordinates (2, -4).
(v) The abscissa of the point D.
(vi) The ordinate of the point H.
(vii)The coordinates of the point L.
(viii) The coordinates of the point M.

 
 
Answer
 

(i) The coordinates of B is (-5, 2).

(ii) The coordinates of C is (5, -5).

(iii) The point identified by the coordinates (-3, -5) is E.

(iv) The point identified by the coordinates (2, -4) is G.

 



(v) Abscissa means x coordinate of point D. So, abscissa of the point D is 6.

(vi) Ordinate means y coordinate of point H. So, ordinate of point H is -3.

(vii) The coordinates of the point L is (0, 5).

(viii) The coordinates of the point M is (- 3, 0).

https://youtu.be/ILVmMPtFim4

Exercise 3.3

1. In which quadrant or on which axis do each of the points (-2, 4), (3, -1), (-1, 0), (1, 2) and (-3, -5) lie? Verify your answer by locating them on the Cartesian plane.

Answer
 
(-2, 4) → Second quadrant
(3, -1) → Fourth quadrant
(-1, 0) → Second quadrant
(1, 2) → First quadrant
(-3, -5) → Third quadrant

2. Plot the points (x, y) given in the following table on the plane, choosing suitable units of distance on the axes.
 
      x         -2               -1               0               1                   3         
y87-1.253-1

Answer

Points (x,y) on the plane. 1unit = 1 cm

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NCERT Solutions for Class 9 Maths Chapter 2 Polynomial | EduGrown

In This Post we are  providing Chapter 2 Polynomials NCERT Solutions for Class 9 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Polynomials Class 9 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 9 Maths Polynomials NCERT Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

https://youtu.be/Zsql4HVImSk

Exercise 2.1
 

1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.

(i) 4x2 – 3x + 7
(ii) y2 + √2

(iii) 3√t + t√2

(iv) y + 2/y
(v) x10 + y3 + t50

Answer
 
(i) 4x2 – 3x + 7
There is only one variable x with whole number power so this polynomial in one variable.
 
(ii)  y2 + √2

There is only one variable y with whole number power so this polynomial in one variable.

 
(iii) 3√2 + t√2 

There is only one variable t but in 3√t power of t is 1/2 which is not a whole number so 3√t + t√2 is not a polynomial.

 
(iv) y + 2/y

There is only one variable y but 2/y = 2y-1 so the power is not a whole number so y + 2/y is not a polynomial.

 
(v) x10 + y3 + t50

There are three variable x, y and t and there powers are whole number so this polynomial in three variable.

 

2. Write the coefficients of x2 in each of the following:
(i) 2 + x2 + x
(ii) 2 – x2 + x3

(iv) √2x – 1

Answer

(i) coefficients of x2 = 1
(ii) coefficients of x2 = -1
(iii) coefficients of x2 = π/2
(iv) coefficients of x2 = 0

3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.

 
Answer
 
3x35+7 and 4x100

4. Write the degree of each of the following polynomials:
(i) 5x3 + 4x2 + 7x 
(ii) 4 – y2 
(iii) 5t – √7
(iv) 3
 
Answer

(i) 5x3 has highest power in the given polynomial which power is 3. Therefore, degree of polynomial is 3.

(ii) – y2  has highest power in the given polynomial which power is 2. Therefore, degree of polynomial is 2.

(iii) 5t has highest power in the given polynomial which power is 1. Therefore, degree of polynomial is 1.

(iv) There is no variable in the given polynomial. Therefore, degree of polynomial is 0.

5. Classify the following as linear, quadratic and cubic polynomial:
(i) x2 + x
► Quadratic Polynomial


(ii) x – x3
► Cubic Polynomial


(iii) y + y2 +4
► Quadratic Polynomial

(iv) 1 + x
► Linear Polynomial

(v) 3t
►Linear Polynomial

(vi) r2
► Quadratic Polynomial

(vii) 7x3
► Cubic Polynomial

https://youtu.be/aZ-LJird–k

Exercise 2.2

1. Find the value of the polynomial at 5x + 4x2 + 3 at

(i) x = 0 (ii) x = – 1 (iii) x = 2
 
Answer
 
(i) p(x) = 5x + 4x2 + 3
    p(0) = 5(0) + 4(0)2 + 3
           = 3

(ii) p(x) = 5x + 4x2 + 3
    p(-1) = 5(-1) + 4(-1)2 + 3
           = -5 + 4(1) + 3 = 2

(iii) p(x) = 5x + 4x2 + 3
    p(2) = 5(2) + 4(2)2 + 3
           = 10 + 16 + 3 = 29

2. Find p(0), p(1) and p(2) for each of the following polynomials:

(i) p(y) = y2 – y + 1

(ii) p(t) = 2 + t + 2t2 – t3
(iii) p(x) = x3 
(iv) p(x) = (x – 1) (x + 1)
 
Answer
 

(i) p(y) = y2 – y + 1
p(0) = (0)2 – (0) + 1 = 1
p(1) = (1)2 – (1) + 1 = 1
p(2) = (2)2 – (2) + 1 = 3

(ii) p(t) = 2 + t + 2t2 – t3
p(0) = 2 + 0 + 2 (0)2 – (0)3 = 2
p(1) = 2 + (1) + 2(1)2 – (1)3
= 2 + 1 + 2 – 1 = 4
p(2) = 2 + 2 + 2(2)2 – (2)3
= 2 + 2 + 8 – 8 = 4

(iii) p(x) = x3
p(0) = (0)3 = 0
p(1) = (1)3 = 1
p(2) = (2)3 = 8


(iv) p(x) = (x – 1) (x + 1)
p(0) = (0 – 1) (0 + 1) = (- 1) (1) = – 1
p(1) = (1 – 1) (1 + 1) = 0 (2) = 0
p(2) = (2 – 1 ) (2 + 1) = 1(3) = 3

Page No: 35

3. Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1, x = -1/3
(ii)  p(x) = 5x – π, x = 4/5

(iii) p(x) = x2 – 1, x = 1, -1

(iv) p(x) = (x + 1) (x – 2), x = -1, 2
(v) p(x) = x2 , x = 0

(viii) p(x) = 2x + 1, x = 1/2

Answer

(i) If x = -1/3 is a zero of polynomial p(x) = 3x + 1 then p(-1/3) should be 0.
At, p(-1/3) = 3(-1/3) + 1 = -1 + 1 = 0
Therefore, x = -1/3 is a zero of polynomial p(x) = 3x + 1.

(ii) If x = 4/5 is a zero of polynomial p(x) = 5x – π then p(4/5) should be 0.
At, p(4/5) = 5(4/5) – π = 4 – π
Therefore, x = 4/5 is not a zero of given polynomial p(x) = 5x – π.

(iii) If x = 1 and x = -1 are zeroes of polynomial p(x) = x2 – 1, then p(1) and p(-1) should be 0.
At, p(1) = (1)2 – 1 = 0 and
At, p(-1) = (-1)2 – 1 = 0
Hence, x = 1 and -1 are zeroes of the polynomial  p(x) = x2 – 1.

(iv) If x = -1 and x = 2 are zeroes of polynomial p(x) = (x +1) (x – 2), then p( – 1) and (2)should be 0.
At, p(-1) = (-1 + 1) (-1 – 2) = 0 (-3) = 0, and
At, p(2) = (2 + 1) (2 – 2) = 3 (0) = 0
Therefore, x = -1 and x = 2 are zeroes of the polynomial p(x) = (x +1) (x – 2).
 
(v) If x = 0 is a zero of polynomial p(x) = x2, then p(0) should be zero.
Here, p(0) = (0)2 = 0
Hence, x = 0 is a zero of the polynomial p(x) = x2.



(viii)  If x = 1/2 is a zero of polynomial p(x) = 2x + 1 then p(1/2) should be 0.
At, p(1/2) = 2(1/2) + 1 = 1 + 1 = 2
Therefore, x = 1/2 is not a zero of given polynomial p(x) = 2x + 1.

4. Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5 
(ii) p(x) = x – 5 
(iii) p(x) = 2x + 5
(iv) p(x) = 3x – 2 
(v) p(x) = 3x 
(vi) p(x) = ax, a ≠ 0
(vii) p(x) = cx + d, c ≠ 0, c, are real numbers.
 
Answer
 
(i) p(x) = x + 5 
p(x) = 0
x + 5 = 0
x = -5
Therefore, x = -5 is a zero of polynomial p(x) = x + 5 .
 
(ii) p(x) = x – 5
p(x) = 0
x – 5 = 0
x = 5
Therefore, x = 5 is a zero of polynomial p(x) = x – 5.
 
(iii) p(x) = 2x + 5
p(x) = 0
2x + 5 = 0
2x = -5
x = -5/2
Therefore, x = -5/2 is a zero of polynomial p(x) = 2x + 5.

(iv) p(x) = 3x – 2
p(x) = 0
3x – 2 = 0
x = 2/3
Therefore, x = 2/3 is a zero of polynomial p(x) = 3x – 2.

(v) p(x) = 3x
p(x) = 0
3x = 0
x = 0
Therefore, x = 0 is a zero of polynomial p(x) = 3x.

(vi) p(x) = ax
p(x) = 0
ax = 0
x = 0
Therefore, x = 0 is a zero of polynomial p(x) = ax.

(vii) p(x) = cx + d
p(x) = 0
cx + d = 0
x = –d/c
Therefore, x = –d/c is a zero of polynomial p(x) = cx + d.

https://youtu.be/sIqZwHl8qw8

Exercises 2.3

1. Find the remainder when x3 + 3x2 + 3x + 1 is divided by
(i) x + 1
(ii) x – 1/2
(iii) x
(iv) x + π
(v) 5 + 2x

Answer

(i) x + 1
By long division,

Therefore, the remainder is 0.

(ii) x – 1/2
By long division,


Therefore, the remainder is 27/8.

(iii) x

Therefore, the remainder is 1.

(iv) x + π

Therefore, the remainder is [1 – 3π + 3π2 – π3].

(v) 5 + 2x

Therefore, the remainder is -27/8.

2. Find the remainder when x3 – ax2 + 6x – a is divided by x – a.

Answer

By Long Division,

Therefore, remainder obtained is 5a when x3 – ax2 + 6x – a is divided by x – a.

3. Check whether 7 + 3x is a factor of 3x3 + 7x.

Answer

We have to divide 3x3 + 7x by 7 + 3x. If remainder comes out to be 0 then 7 + 3x will be a factor of 3x3 + 7x.
By Long Division,

As remainder is not zero so 7 + 3x is not a factor of 3x3 + 7x.

https://youtu.be/IXQ8DZKnyI0

Exercise 2.4

1. Determine which of the following polynomials has (x + 1) a factor:
(i) x3 + x2 + x + 1

(ii) x4 + x3 + x2 + x + 1
(iii) x4 + 3x3 + 3x2 + x + 1 
(iv) x3 – x2 – (2 + √2)x + √2

Answer

(i) If (x + 1) is a factor of p(x) = x3 + x2 + x + 1, p(-1) must be zero. 
Here, p(x) = x3 + x2 + x + 1 
p(-1) = (-1)3 + (-1)2 + (-1) + 1 
= -1 + 1 – 1 + 1 = 0
Therefore, x + 1 is a factor of this polynomial

(ii) If (x + 1) is a factor of p(x) = x4 + x3 + x2 + x + 1, p(-1) must be zero. 
Here, p(x) = x4 + x3 + x2 + x + 1 
p(-1) = (-1)4 + (-1)3 + (-1)2 + (-1) + 1
= 1 – 1 + 1 – 1 + 1 = 1

As, p(-1) ≠ 0
Therefore, x + 1 is not a factor of this polynomial

(iii)If (x + 1) is a factor of polynomial p(x) = x4 + 3x3 + 3x2 + x + 1, p(- 1) must be 0. 
p(-1) = (-1)4 + 3(-1)3 + 3(-1)2 + (-1) + 1
= 1 – 3 + 3 – 1 + 1 = 1
As, p(-1) ≠ 0
Therefore, x + 1 is not a factor of this polynomial.

 

(iv) If (x + 1) is a factor of polynomial

p(x) = x3 – x2 – (2 + √2)x + √2, p(- 1) must be 0.

p(-1) =  (-1)3 –  (-1)2 –  (2 + √2) (-1) + √2
= -1 – 1 + 2 + √2 + √2
=2√2
As, p(-1) ≠ 0
Therefore,, x + 1 is not a factor of this polynomial.

2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i) p(x) = 2x3 + x2 – 2x – 1, g(x) = x + 1
(ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2
(iii) p(x) = x3 – 4 x2 + x + 6, g(x) = x – 3

Answer

(i) If g(x) = x + 1 is a factor of given polynomial p(x), p(- 1) must be zero.
p(x) = 2x3 + x2 – 2x – 1
p(- 1) = 2(-1)3 + (-1)2 – 2(-1) – 1
= 2(- 1) + 1 + 2 – 1 = 0
Hence, g(x) = x + 1 is a factor of given polynomial.

(ii) If g(x) = x + 2 is a factor of given polynomial p(x), p(- 2) must be 0.
p(x) = x3 +3x2 + 3x + 1
p(-2) = (-2)3 + 3(- 2)2 + 3(- 2) + 1
= -8 + 12 – 6 + 1
= -1

As, p(-2) ≠ 0
Hence g(x) = x + 2 is not a factor of given polynomial.


(iii) If g(x) = x – 3 is a factor of given polynomial p(x), p(3) must be 0.
p(x) = x3 – 4x2 + x + 6
p(3) = (3)3 – 4(3)2 + 3 + 6
= 27 – 36 + 9 = 0
Therefore,, g(x) = x – 3 is a factor of given polynomial.

Page No: 44

3. Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:(i) p(x) = x2 + x + k
(ii) p(x) = 2x2 + kx +  √2
(iii) p(x) = kx2 – √2x + 1
(iv) p(x) = kx2 – 3x + k

Answer

(i) If x – 1 is a factor of polynomial p(x) = x2 + x + k, then

p(1) = 0
⇒ (1)2 + 1 + k = 0
⇒ 2 + k = 0
⇒ k = – 2
Therefore, value of k is -2.


(ii) If x – 1 is a factor of polynomial p(x) = 2x2 + kx +  √2, then
p(1) = 0
⇒ 2(1)2 + k(1) + √2 = 0
⇒ 2 + k + √2 = 0
⇒ k = -2 – √2 = -(2 + √2)

Therefore, value of k is -(2 + √2).


(iii) If x – 1 is a factor of polynomial p(x) = kx2 – √2x + 1, then
p(1) = 0
⇒ k(1)2 – √2(1) + 1 = 0
⇒ k – √2 + 1 = 0
⇒ k = √2 – 1

Therefore, value of k is √2 – 1.


(iv) If x – 1 is a factor of polynomial p(x) = kx2 – 3x + k, then
p(1) = 0
⇒ k(1)2 + 3(1) + k = 0
⇒ k – 3 + k = 0
⇒ 2k – 3 = 0
⇒ k = 3/2

Therefore, value of k is 3/2.


4. Factorise:
(i) 12x2 + 7x + 1
(ii) 2x2 + 7x + 3
(iii) 6x2 + 5x – 6
(iv) 3x2 – x – 4 

Answer

(i) 12x2 + 7x + 1
= 12x2 – 4x – 3x+ 1                   
= 4x (3x – 1) – 1 (3x – 1)
= (3x – 1) (4x – 1)

(ii) 2x2 + 7x + 3
= 2x2 + 6x + x + 3
= 2x (x + 3) + 1 (x + 3)
=  (x + 3) (2x + 1) 

(iii) 6x2 + 5x – 6
= 6x2 + 9x – 4x – 6

 = 3x (2x + 3) – 2 (2x + 3)
= (2x + 3) (3x – 2)


(iv) 3x2 – x – 4
= 3x2 – 4x + 3x – 4 
= x (3x – 4) + 1 (3x – 4)
= (3x – 4) (x + 1)

5. Factorise:
(i) x3 – 2x2 – x + 2

(ii) x3 – 3x2 – 9x – 5
(iii) x3 + 13x2 + 32x + 20 
(iv) 2y3 + y2 – 2y – 1

Answer

(i) Let p(x) = x3 – 2x2 – x + 2
Factors of 2 are ±1 and ± 2
By trial method, we find that
p(1) = 0
So, (x+1) is factor of p(x)
Now,
p(x) = x3 – 2x2 – x + 2
p(-1) = (-1)3 – 2(-1)2 – (-1) + 2 = -1 -2 + 1 + 2 = 0
Therefore, (x+1) is the factor of  p(x)

 
Now, Dividend = Divisor × Quotient + Remainder
(x+1) (x2 – 3x + 2)
= (x+1) (x2 – x – 2x + 2)
= (x+1) {x(x-1) -2(x-1)}
= (x+1) (x-1) (x+2)


(ii) Let p(x) = x3 – 3x2 – 9x – 5
Factors of 5 are ±1 and ±5
By trial method, we find that
p(5) = 0
So, (x-5) is factor of p(x)
Now,
p(x) = x3 – 2x2 – x + 2
p(5) = (5)3 – 3(5)2 – 9(5) – 5 = 125 – 75 – 45 – 5 = 0
Therefore, (x-5) is the factor of  p(x)

Now, Dividend = Divisor × Quotient + Remainder

(x-5) (x2 + 2x + 1)

= (x-5) (x2 + x + x + 1)
= (x-5) {x(x+1) +1(x+1)}
= (x-5) (x+1) (x+1)

(iii) Let p(x) = x3 + 13x2 + 32x + 20
Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20
By trial method, we find that
p(-1) = 0
So, (x+1) is factor of p(x)
Now,
p(x) =  x3 + 13x2 + 32x + 20
p(-1) = (-1)3 + 13(-1)2 + 32(-1) + 20 = -1 + 13 – 32 + 20 = 0
Therefore, (x+1) is the factor of  p(x)

 
Now, Dividend = Divisor × Quotient + Remainder
(x+1) (x2 + 12x + 20)
= (x+1) (x2 + 2x + 10x + 20)
= (x-5) {x(x+2) +10(x+2)}
= (x-5) (x+2) (x+10)

(iv) Let p(y) = 2y3 + y2 – 2y – 1
Factors of ab = 2× (-1) = -2 are ±1 and ±2
By trial method, we find that
p(1) = 0
So, (y-1) is factor of p(y)
Now,
p(y) =  2y3 + y2 – 2y – 1
p(1) = 2(1)3 + (1)2 – 2(1) – 1 = 2 +1 – 2 – 1 = 0
Therefore, (y-1) is the factor of  p(y)

 
 Now, Dividend = Divisor × Quotient + Remainder
(y-1) (2y2 + 3y + 1)
= (y-1) (2y2 + 2y + y + 1)
= (y-1) {2y(y+1) +1(y+1)}
= (y-1) (2y+1) (y+1)

Part-1 ( Q1 to Q12)

https://youtu.be/eUwEQgj6dgU

Part-2 ( Q13 to Q16)

https://youtu.be/1qa2EdkkBGE

Exercise 2.5

1. Use suitable identities to find the following products:
    (i) (x + 4) (x + 10)                     (ii) (x + 8) (x – 10)                      (iii) (3x + 4) (3x – 5)
    (iv) (y2 + 3/2) (y2 – 3/2)             (v) (3 – 2x) (3 + 2x)

Answer

(i) Using identity, (x + a) (x + b) = x2 + (a + b) x + ab 
In (x + 4) (x + 10), a = 4 and b = 10
Now,
(x + 4) (x + 10) = x2 + (4 + 10)x + (4 × 10)
                         = x2 + 14x + 40

(ii) (x + 8) (x – 10)
Using identity, (x + a) (x + b) = x2 + (a + b) x + ab
Here, a = 8 and b = –10
(x + 8) (x – 10) = x2 + {8 +(– 10)}x + {8×(– 10)}
                         = x2 + (8 – 10)x – 80
                         = x2 – 2x – 80

(iii) (3x + 4) (3x – 5)
Using identity, (x + a) (x + b) = x2 + (a + b) x + ab
Here, x = 3x , a = 4 and b = -5
(3x + 4) (3x – 5) = (3x) 2 + {4 + (-5)}3x + {4×(-5)}
                           = 9x2 + 3x(4 – 5) – 20
                           = 9x2 – 3x – 20

(iv) (y2 + 3/2) (y2 – 3/2)
Using identity, (x + y) (x –y) = x2 – y2
Here, x = y2 and y = 3/2
(y2 + 3/2) (y2 – 3/2) = (y2)2 – (3/2)2
                                         = y4 – 9/4

(v) (3 – 2x) (3 + 2x)
Using identity, (x + y) (x –y) = x2 – y2
Here, x = 3 and y = 2x
(3 – 2x) (3 + 2x) = 32 – (2x)2
                                   =  9 – 4x2

2. Evaluate the following products without multiplying directly:
    (i) 103 × 107               (ii) 95 × 96               (iii) 104 × 96


Answer

(i) 103 × 107 = (100 + 3) (100 + 7)
Using identity, (x + a) (x + b) = x2 + (a + b) x + ab
Here, x = 100, a = 3 and b = 7
103 × 107 = (100 + 3) (100 + 7) = (100)2 + (3 + 7)10 + (3 × 7)
                 = 10000 + 100 + 21 
                 = 10121

(ii) 95 × 96 = (90 + 5) (90 + 4)
Using identity, (x + a) (x + b) = x2 + (a + b) x + ab 
Here, x = 90, a = 5 and b = 4
95 × 96 = (90 + 5) (90 + 4) = 902 + 90(5 + 6) + (5 × 6) 
 
             = 8100 + (11 × 90) + 30
             = 8100 + 990 + 30 = 9120
(iii) 104 × 96 = (100 + 4) (100 – 4)
Using identity, (x + y) (x –y) = x2 – y2
Here, x = 100 and y = 4
104 × 96 = (100 + 4) (100 – 4) = (100)2 – (4)2 = 10000 – 16 = 9984 

3. Factorise the following using appropriate identities:
   (i) 9x2 + 6xy + y2                 (ii) 4y2 – 4y + 1              (iii) x2 – y2/100

Answer

(i) 9x2 + 6xy + y2  = (3x) 2 + (2×3x×y) + y2
Using identity, (a + b)2 = a2 + 2ab + b2
Here, a = 3x and b = y
9x2 + 6xy + y2  = (3x) 2 + (2×3x×y) + y2 = (3x + y)2 = (3x + y) (3x + y)

(ii) 4y2 – 4y + 1 = (2y)2 – (2×2y×1) + 1
Using identity, (a – b)2 = a2 – 2ab + b2
Here, a = 2y and b = 1
4y2 – 4y + 1 = (2y)2 – (2×2y×1) + 12 = (2y – 1)2 = (2y – 1) (2y – 1)

(iii) x2 – y2/100 = x2 – (y/10)2
Using identity, a2 – b2 = (a + b) (a – b)
Here, a = x and b = (y/10)
x2 – y2/100 = x2 – (y/10)2 = (x – y/10) (x + y/10)

Page No: 49

4. Expand each of the following, using suitable identities:
    (i) (x + 2y + 4z)2                     (ii) (2x – y + z)2                    (iii) (–2x + 3y + 2z)2
    (iv) (3a – 7b – c)2                         (v) (–2x + 5y – 3z)2                   (vi) [1/4 a – 1/2 b + 1]2  

Answer

(i) (x + 2y + 4z)2
Using identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca  
Here, a = x, b = 2y and c = 4z
(x + 2y + 4z)2 = x2 + (2y)2 + (4z)2 + (2×x×2y) + (2×2y×4z) + (2×4z×x)
                      = x2 + 4y2 + 16z2 + 4xy + 16yz + 8xz

(ii)  (2x – y + z)2
Using identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca  
Here, a = 2x, b = –y and c = z
(2x – y + z)2 = (2x)2 + (-y)2 + z2 + (2×2x×-y) + (2×-y×z) + (2×z×2x) 
                     = 4x2 + y2 + z2 – 4xy – 2yz + 4xz

(iii) (–2x + 3y + 2z)2 
Using identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca  
Here, a = -2x, b = 3y and c = 2z
(–2x + 3y + 2z)2 = (-2x)2 + (3y)2 + (2z)2 + (2×-2x×3y) + (2×3y×2z) + (2×2z×-2x) 
                     = 4x2 + 9y2 + 4z2 – 12xy + 12yz – 8xz

(iv) (3a – 7b – c)2
Using identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca  
Here, a = 3a, b = -7b and c = -c
(3a – 7b – c)2 = (3a)2 + (-7b)2 + (-c)2 + (2×3a×-7b) + (2×-7b×-c) + (2×-c×3a) 
                     = 9a2 + 49b2 + c2 – 42ab + 14bc – 6ac

(v) (–2x + 5y – 3z)2  
Using identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca  
Here, a = -2x, b = 5y and c = -3z
(–2x + 5y – 3z)2 = (-2x)2 + (5y)2 + (-3z)2 + (2×-2x×5y) + (2×5y×-3z) + (2×-3z×-2x) 
                     = 4x2 + 25y2 + 9z2 – 20xy – 30yz + 12xz

(vi) [1/4 a – 1/2 b + 1]2 
Using identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca  
Here, a = 1/4 a, b = -1/2 b and c = 1
[1/4 a – 1/2 b + 1]2 = (1/4 a)2 + (-1/2 b)2 + 12 + (2×1/4 a×-1/2 b) + (2×-1/2 b×1) + (2×1×1/4 a) 
                                = 1/16 a2 + 1/4 b2 + 1 – 1/4 ab – b + 1/2 a

5. Factorise:
(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
(ii) 2x2 + y2 + 8z2 – 2√2 xy + 4√2 yz – 8xz

Answer

(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
Using identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca 
4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
= (2x)2 + (3y)2 + (-4z)2 + (2×2x×3y) + (2×3y×-4z) + (2×-4z×2x)
= (2x + 3y – 4z)2
=  (2x + 3y – 4z) (2x + 3y – 4z)

(ii) 2x2 + y2 + 8z2 – 2√2 xy + 4√2 yz – 8xz
Using identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
2x2 + y2 + 8z2 – 2√2 xy + 4√2 yz – 8xz 
= (-√2x)2 + (y)2 + (2√2z)2 + (2×-√2x×y) + (2×y×2√2z) + (2×2√2z×-√2x)
= (-√2x + y + 2√2z)2
=  (-√2x + y + 2√2z) (-√2x + y + 2√2z)

6. Write the following cubes in expanded form:
    (i) (2x + 1)3                 (ii) (2a – 3b)3                (iii) [3/2 x + 1]3           (iv) [x – 2/3 y]3 

Answer

(i) (2x + 1)3
Using identity, (a + b)3 = a3 + b3 + 3ab(a + b)
(2x + 1)3 = (2x)3 + 13 + (3×2x×1)(2x + 1)
= 8x3 + 1 + 6x(2x + 1)
= 8x3 + 12x2 + 6x + 1

(ii) (2a – 3b)3
Using identity, (a – b)3 = a3 – b3 – 3ab(a – b) 
(2a – 3b)3 = (2a)3 – (3b)3 – (3×2a×3b)(2a – 3b)
= 8a3 – 27b3 – 18ab(2a – 3b)
= 8a3 – 27b3 – 36a2b + 54ab2

(iii) [3/2 x + 1]3 
Using identity, (a + b)3 = a3 + b3 + 3ab(a + b)
[3/2 x + 1]3 = (3/2 x)3 + 13 + (3×3/2 x×1)(3/2 x + 1)
= 27/8 x3 + 1 + 9/2 x(3/2 x + 1)
= 27/8 x3 + 1 + 27/4 x2 + 9/2 x
= 27/8 x3 + 27/4 x2 + 9/2 x + 1

(iv) [x – 2/3 y]3
Using identity, (a – b)3 = a3 – b3 – 3ab(a – b)
[x – 2/3 y]3 = (x)3 – (2/3 y)3 – (3×x×2/3 y)(x – 2/3 y)
= x3 – 8/27y3 – 2xy(x – 2/3 y)
= x3 – 8/27y3 – 2x2y + 4/3xy2

7. Evaluate the following using suitable identities: 
    (i) (99)3            (ii) (102)3             (iii) (998)3  

Answer

(i) (99)3 = (100 – 1)3
Using identity, (a – b)3 = a3 – b3 – 3ab(a – b) 
(100 – 1)3 = (100)3 – 13 – (3×100×1)(100 – 1)
= 1000000 – 1 – 300(100 – 1)
= 1000000 – 1 – 30000 + 300
= 970299

(ii) (102)3 = (100 + 2)3
Using identity, (a + b)3 = a3 + b3 + 3ab(a + b)
(100 + 2)3 = (100)3 + 23 + (3×100×2)(100 + 2)
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200
= 1061208

(iii) (998)3 
Using identity, (a – b)3 = a3 – b3 – 3ab(a – b) 
(1000 – 2)3 = (1000)3 – 23 – (3×1000×2)(1000 – 2)
= 100000000 – 8 – 6000(1000 – 2)
= 100000000 – 8- 600000 + 12000
= 994011992

8. Factorise each of the following: 
(i) 8a3 + b3 + 12a2b + 6ab2                           (ii) 8a3 – b3 – 12a2b + 6ab2
(iii) 27 – 125a3 – 135a + 225a2                      (iv) 64a3 – 27b3 – 144a2b + 108ab2
(v) 27p3 – 1/216 – 9/2 p2 + 1/4 p

Answer


(i) 8a3 + b3 + 12a2b + 6ab2
Using identity, (a + b)3 = a3 + b3 + 3a2b + 3ab2
8a3 + b3 + 12a2b + 6ab2 
= (2a)3 + b3 + 3(2a)2b + 3(2a)(b)2
= (2a + b)3
= (2a + b)(2a + b)(2a + b)

(ii) 8a3 – b3 – 12a2b + 6ab2
Using identity, (a – b)3 = a3 – b3 – 3a2b + 3ab2
8a3 – b3 – 12a2b + 6ab2= (2a)3 – b3 – 3(2a)2b + 3(2a)(b)2
= (2a – b)3
= (2a – b)(2a – b)(2a – b)

(iii) 27 – 125a3 – 135a + 225a2
Using identity, (a – b)3 = a3 – b3 – 3a2b + 3ab2
27 – 125a3 – 135a + 225a2= 33 – (5a)3 – 3(3)2(5a) + 3(3)(5a)2
= (3 – 5a)3
= (3 – 5a)(3 – 5a)(3 – 5a)

(iv) 64a3 – 27b3 – 144a2b + 108ab2
Using identity, (a – b)3 = a3 – b3 – 3a2b + 3ab2
64a3 – 27b3 – 144a2b + 108ab2= (4a)3 – (3b)3 – 3(4a)2(3b) + 3(4a)(3b)2
= (4a – 3b)3
= (4a – 3b)(4a – 3b)(4a – 3b)

(v) 27p3 – 1/216 – 9/2 p2 + 1/4 p 

 Using identity, (a – b)3 = a3 – b3 – 3a2b + 3ab2
 27p3 – 1/216 – 9/2 p2 + 1/4 p
= (3p)3 – (1/6)3 – 3(3p)2(1/6) + 3(3p)(1/6)2
= (3p – 1/6)3
= (3p – 1/6)(3p – 1/6)(3p – 1/6)

9. Verify : (i) x3 + y3 = (x + y) (x2 – xy + y2)             (ii) x3 – y3 = (x – y) (x2 + xy + y2)

Answer

(i) x3 + y3 = (x + y) (x2 – xy + y2)
We know that, 
(x + y)3 = x3 + y3 + 3xy(x + y) 
⇒ x3 + y3 = (x + y)3 – 3xy(x + y)
⇒ x3 + y3 = (x + y)[(x + y)2 – 3xy]                  {Taking (x+y) common}
⇒ x3 + y3 = (x + y)[(x2 + y2 + 2xy) – 3xy] 
⇒ x3 + y3 = (x + y)(x2 + y2 – xy) 

(ii) x3 – y3 = (x – y) (x2 + xy + y2 )
We know that, 
(x – y)3 = x3 – y3 – 3xy(x – y) 
⇒ x3 – y3 = (x – y)3 + 3xy(x – y)
⇒ x3 + y3 = (x – y)[(x – y)2 + 3xy]                     {Taking (x-y) common}
⇒ x3 + y3 = (x – y)[(x2 + y2 – 2xy) + 3xy] 
⇒ x3 + y3 = (x + y)(x2 + y2 + xy)

10. Factorise each of the following:
      (i) 27y3 + 125z3                     (ii) 64m3 – 343n3

Answer

(i) 27y3 + 125z3
Using identity, x3 + y3 = (x + y) (x2 – xy + y2)
27y3 + 125z3 = (3y)3 + (5z)3
= (3y + 5z) {(3y)2 – (3y)(5z) + (5z)2}
= (3y + 5z) (9y2 – 15yz + 25z)2 

(ii) 64m3 – 343n3 
Using identity, x3 – y3 = (x – y) (x2 + xy + y2 ) 
64m3 – 343n3 = (4m)3 – (7n)3
= (4m + 7n) {(4m)2 + (4m)(7n) + (7n)2}
= (4m + 7n) (16m2 + 28mn + 49n)2 

11. Factorise : 27x3 + y3 + z3 – 9xyz

Answer


27x3 + y3 + z3 – 9xyz = (3x)3 + y3 + z3 – 3×3xyz
Using identity, x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – xz)
27x3 + y3 + z3 – 9xyz
= (3x + y + z) {(3x)2 + y2 + z2 – 3xy – yz – 3xz}
= (3x + y + z) (9x2 + y2 + z2 – 3xy – yz – 3xz)

12. Verify that: x3 + y3 + z3 – 3xyz = 1/2(x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]

Answer

We know that,
x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – xz)
⇒ x3 + y3 + z3 – 3xyz = 1/2×(x + y + z) 2(x2 + y2 + z2 – xy – yz – xz)

= 1/2(x + y + z) (2x2 + 2y2 + 2z2 – 2xy – 2yz – 2xz) 
= 1/2(x + y + z) [(x2 + y2 -2xy) + (y2 + z2 – 2yz) + (x2 + z2 – 2xz)] 
= 1/2(x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]

13. If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.

Answer

We know that,
x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – xz)
Now put (x + y + z) = 0,  
x3 + y3 + z3 – 3xyz = (0)(x2 + y2 + z2 – xy – yz – xz) 
⇒ x3 + y3 + z3 – 3xyz = 0


14. Without actually calculating the cubes, find the value of each of the following:
     (i) (-12)3 + (7)3 + (5)3
     (ii) (28)3 + (–15)3 + (-13)3

Answer

(i) (-12)3 + (7)3 + (5)3
 Let x = -12, y = 7 and z = 5
We observed that, x + y + z = -12 + 7 + 5 = 0

We know that if,
x + y + z = 0, then x3 + y3 + z3 = 3xyz
(-12)3 + (7)3 + (5)3 = 3(-12)(7)(5) = -1260

(ii) (28)3 + (–15)3 + (-13)3
 Let x = 28, y = -15 and z = -13
We observed that, x + y + z = 28 – 15 – 13 = 0

We know that if,
x + y + z = 0, then x3 + y3 + z3 = 3xyz
(28)3 + (–15)3 + (-13)3 = 3(28)(-15)(-13) = 16380

15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: 
(i) Area : 25a2 – 35a + 12
(ii) Area : 35 y2 + 13y – 12

Answer


(i) Area : 25a2 – 35a + 12

Since, area is product of length and breadth therefore by factorizing the given area, we can know the length and breadth of rectangle.
25a2 – 35a + 12
= 25a2 – 15a -20a + 12
= 5a(5a – 3) – 4(5a – 3)
= (5a – 4)(5a – 3)
Possible expression for length = 5a – 4
Possible expression for breadth = 5a – 3

(ii) Area : 35 y2 + 13y – 12
35 y2 + 13y – 12
= 35y2 – 15y + 28y – 12
= 5y(7y – 3) + 4(7y – 3)
= (5y + 4)(7y – 3)
Possible expression for length = (5y + 4)
Possible expression for breadth = (7y – 3)

Page No: 50

16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (i) Volume : 3x2 – 12x
(ii) Volume : 12ky2 + 8ky – 20k

Answer

(i) Volume : 3x2 – 12x 
Since, volume is product of length, breadth and height therefore by factorizing the given volume, we can know the length, breadth and height of the cuboid. 
3x2 – 12x
= 3x(x – 4)
Possible expression for length = 3
Possible expression for breadth = x
Possible expression for height = (x – 4)

(ii) Volume : 12ky2 + 8ky – 20k 
Since, volume is product of length, breadth and height therefore by factorizing the given volume, we can know the length, breadth and height of the cuboid. 
12ky2 + 8ky – 20k
= 4k(3y2 + 2y – 5)
= 4k(3y2 +5y – 3y – 5)
= 4k[y(3y +5) – 1(3y + 5)]
= 4k (3y +5) (y – 1)
Possible expression for length = 4k
Possible expression for breadth = (3y +5)
Possible expression for height = (y – 1)

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NCERT Solutions for Class 9 Maths Chapter 1 Number System | EduGrown

In This Post we are  providing Chapter 1 Number System NCERT Solutions for Class 9 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Number system Class 9 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 9 Maths Number System NCERT Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

NCERT Solutions for Class 9 Maths Chapter 1 Number System

https://youtu.be/JRB8Nhg_OMY

Exercise 1.1

1. Is zero a rational number? Can you write it in the form p/q, where p and q are integers and q ≠ 0?

 
Answer
 

Yes. Zero is a rational number as it can be represented as 0/1 or 0/2.

2. Find six rational numbers between 3 and 4.

Answer

There are infinite rational numbers in between 3 and 4.
3 and 4 can be represented as 24/8 and 32/8 respectively.

Therefore, six rational numbers between 3 and 4 are
25/8, 26/8, 27/8, 28/8, 29/8, 30/8.

3. Find five rational numbers between 3/5 and 4/5.

Answer

There are infinite rational numbers in between 3/5 and 4/5
3/5 = 3×6/5×6 = 18/30
4/5 = 4×6/5×6 = 24/30
Therefore, five rational numbers between 3/5 and 4/5 are
19/30, 20/30, 21/30, 22/30, 23/30.

4. State whether the following statements are true or false. Give reasons for your answers.

(i) Every natural number is a whole number.
► True, since the collection of whole numbers contains all natural numbers.
 
(ii) Every integer is a whole number.
► False, as integers may be negative but whole numbers are always positive.

(iii) Every rational number is a whole number.
► False, as rational numbers may be fractional but whole numbers may not be.

https://youtu.be/WwvqdJvSWvg

Exercise 1.2

1. State whether the following statements are true or false. Justify your answers.

(i) Every irrational number is a real number.

► True, since the collection of real numbers is made up of rational and irrational numbers.

(ii) Every point on the number line is of the form√m, where m is a natural number.
► False, since positive number cannot be expressed as square roots.

(iii) Every real number is an irrational number.
► False, as real numbers include both rational and irrational numbers. Therefore, every real number cannot be an irrational number.

2. Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.


Answer

No, the square roots of all positive integers are not irrational. For example √4 = 2.

3. Show how √5 can be represented on the number line.

Answer

Step 1: Let AB be a line of length 2 unit on number line.
Step 2: At B, draw a perpendicular line BC of length 1 unit. Join CA.

Step 3: Now, ABC is a right angled triangle. Applying Pythagoras theorem,
AB2 + BC2 = CA2
⇒ 22 + 12 = CA2
⇒ CA2 = 5
⇒ CA = √5
Thus, CA is a line of length √5 unit.
Step 4: Taking CA as a radius and A as a centre draw an arc touching
the number line. The point at which number line get intersected by
arc is at √5 distance from 0 because it is a radius of the circle
whose centre was A.
Thus, √5 is represented on the number line as shown in the figure.
root 5 on the number line

https://youtu.be/15ZUeooo9x0

Exercise 1.3

1. Write the following in decimal form and say what kind of decimal expansion each has:
(i) 36/100
= 0.36 (Terminating)

(ii) 1/11
0.09090909… = 0.9 (Non terminating repeating)

(iii)
= 33/8 = 4.125 (Terminating)

(iv) 3/13
= 0.230769230769… = 0.230769 (Non terminating repeating)

(v) 2/11
= 0.181818181818… = 0.18 (Non terminating repeating)

(vi) 329/400
= 0.8225 (Terminating)

2. You know that 1/7 = 0.142857.Can you predict what the decimal expansion of 2/7, 3/7, 4/7, 5/7, 6/7 are without actually doing the long division? If so, how?
[Hint: Study the remainders while finding the value of 1/7 carefully.]

Answer

Yes. We can be done this by:


3. Express the following in the form p/q where p and q are integers and q ≠ 0.

(i) 0.6
(ii) 0.47
(iii) 0.001
 
Answer
 
(i) 0.6 = 0.666…
Let x = 0.666…
10x = 6.666…
10x = 6 + x
9x = 6
x = 2/3

(ii) 0.47 = 0.4777…
= 4/10 + 0.777/10
Let x = 0.777…
10x = 7.777…
10x = 7 + x
x = 7/9
4/10 + 0.777…/10 = 4/10 + 7/90
= 36/90 +7/90 = 43/90
 
(iii) 0.001 = 0.001001…
Let x = 0.001001…

1000x = 1.001001…
1000x = 1 + x
999x = 1

x = 1/999

4. Express 0.99999…in the form  p/q. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.

Answer

Let x = 0.9999…
10x = 9.9999…
10x = 9 + x
9x = 9
x = 1
The difference between 1 and 0.999999 is 0.000001 which is negligible. Thus, 0.999 is too much near 1, Therefore, the 1 as answer can be justified. 
 

5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17? Perform the division to check your answer.

Answer

1/17 = 0.0588235294117647
There are 16 digits in the repeating block of the decimal expansion of 1/17.
Division Check:

= 0.0588235294117647


6. Look at several examples of rational numbers in the form p/q (q ≠ 0) where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?

Answer

We observe that when q is 2, 4, 5, 8, 10… then the decimal expansion is terminating. For example:
1/2 = 0.5, denominator q = 21
7/8 = 0.875, denominator q = 23
4/5 = 0.8, denominator q = 51

We can observed that terminating decimal may be obtained in the situation where prime factorisation of the denominator of the given fractions has the power of 2 only or 5 only or both.

 

7. Write three numbers whose decimal expansions are non-terminating non-recurring.


Answer
 
Three numbers whose decimal expansions are non-terminating non-recurring are:
0.303003000300003…
0.505005000500005…

0.7207200720007200007200000…

 

8. Find three different irrational numbers between the rational numbers 5/7 and 9/11.

 
Answer
 
5/7 = 0.714285
9/11 = 0.81
Three different irrational numbers are:

0.73073007300073000073…

0.75075007300075000075…
0.76076007600076000076…
 

9. Classify the following numbers as rational or irrational:

(i) √23

(ii) √225

(iii) 0.3796
(iv) 7.478478 
(v) 1.101001000100001…
 
Answer
 
(i) √23 = 4.79583152331…
Since the number is non-terminating non-recurring therefore, it is an irrational number.
 
(ii) √225 = 15 = 15/1
Since the number is rational number as it can represented in p/q form.
 
(iii) 0.3796
Since the number is terminating therefore, it is an rational number.
 
(iv) 7.478478 = 7.478

Since the this number is non-terminating recurring, therefore, it is a rational number.

 
(v) 1.101001000100001…

Since the number is non-terminating non-repeating, therefore, it is an irrational number.

https://youtu.be/qla_5QAJgQ0

Exercises 1.4
 

1. Visualise 3.765 on the number line using successive magnification.

 
Answer
 
3.765 on number line
 
2. Visualize 4.26 on the number line, up to 4 decimal places.
 
Answer
 
4.26 = 4.2626
4.2626 on number line

https://youtu.be/YbI8J1SUzjs

Exercise 1.5
 
1. Classify the following numbers as rational or irrational:
(i) 2 – √5
(ii) (3 + √23) – √23
 
(iii) 2√7/7√7
(iv) 1/√2
(v) 2π
 
Answer
 
(i) 2 – √5 = 2 – 2.2360679… = – 0.2360679…
Since the number is is non-terminating non-recurring therefore, it is an irrational number.

(ii) (3 + √23) – √23 = 3 + √23 – √23 = 3 = 3/1
Since the number is rational number as it can represented in p/q form.

(iii) 2√7/7√7 = 2/7
Since the number is rational number as it can represented in p/q form.

(iv) 1/√2 = √2/2 = 0.7071067811…
Since the number is is non-terminating non-recurring therefore, it is an irrational number.

(v) 2π = 2 × 3.1415… = 6.2830…
Since the number is is non-terminating non-recurring therefore, it is an irrational number.

2. Simplify each of the following expressions:
 
(i) (3 + √3) (2 + √2)
(ii) (3 + √3) (3 – √3)
(iii) (√5 + √2)2
(iv) (√5 – √2) (√5 + √2)
 
Answer
 
(i) (3 + √3) (2 + √2)
⇒ 3 × 2 + 2 + √3 + 3√2+ √3 ×√2
⇒ 6 + 2√3 +3√2 + √6
 
(ii) (3 + √3) (3 – √3) [∵ (a + b) (a – b) = a2 – b2]
⇒ 32 – (√3)2
⇒ 9 – 3
⇒ 6
 
(iii) (√5 + √2)2 [∵ (a + b)2 = a2 + b2 + 2ab]
⇒ (√5)2 + (√2)2 + 2 ×√5 × √2
⇒ 5 + 2 + 2 × √5× 2 
⇒ 7 +2√10

(iv) (√5 – √2) (√5 + √2) [∵ (a + b) (a – b) = a2 – b2]
⇒ (√5)2 – (√2)2 
⇒ 5 – 2
⇒ 3
 
3. Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
 
Answer
 
There is no contradiction. When we measure a value with a scale, we only obtain an approximate value. We never obtain an exact value. Therefore, we may not realise that either c or d is irrational. The value of π is almost equal to 22/7 or 3.142857…
 
4. Represent √9.3 on the number line.
 
Answer

Step 1: Draw a line segment of unit 9.3. Extend it to C so that BC is of 1 unit.
Step 2: Now, AC = 10.3 units. Find the centre of AC and name it as O.
Step 3: Draw a semi circle with radius OC and centre O.
Step 4: Draw a perpendicular line BD to AC at point B which intersect the semicircle at D. Also, Join OD.
Step 5: Now, OBD is a right angled triangle.
Here, OD = 10.3/2 (radius of semi circle), OC = 10.3/2, BC = 1
OB = OC – BC = (10.3/2) – 1 = 8.3/2
Using Pythagoras theorem,
OD2 = BD2 + OB2
⇒ (10.3/2)2 = BD2 + (8.3/2)2
⇒ BD2 = (10.3/2)2 – (8.3/2)2
⇒ BD2 = (10.3/2 – 8.3/2) (10.3/2 + 8.3/2)
⇒ BD2 = 9.3
⇒ BD2 =  √9.3
Thus, the length of BD is √9.3.
Step 6: Taking BD as radius and B as centre draw an arc which touches the line segment. The point where it touches the line segment is at a distance of √9.3 from O as shown in the figure.
root 9.3 on number line
5. Rationalise the denominators of the following:
(i) 1/√7
(ii) 1/√7-√6
(iii) 1/√5+√2
(iv) 1/√7-2
 
Answer

https://youtu.be/Zh_78OJEG38

Exercise 1.6

1. Find:
(i) 641/2
(ii) 321/5
(iii) 1251/3

Answer



2. Find:
(i) 93/2
(ii) 322/5
(iii) 163/4
(iv) 125-1/3
 
Answer



3. Simplify:
(i) 22/3.21/5
(ii) (1/33)7
(iii) 111/2/111/4
(iv) 71/2.81/2

Answer

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NCERT Solutions for Class 10 Maths Chapter 15 Probability | EduGrown

In This Post we are  providing Chapter 15 Probability NCERT Solutions for Class 10 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Probability Class 10 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 10 Maths Probability NCERT Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

NCERT Solutions for Class 10 Maths Chapter 15 Probability

Exercise 15.1

1. Complete the following statements:
(i) Probability of an event E + Probability of the event ‘not E’ = ___________ .
(ii) The probability of an event that cannot happen is __________. Such an event is called ________ .
(iii) The probability of an event that is certain to happen is _________ . Such an event is called _________ .
(iv) The sum of the probabilities of all the elementary events of an experiment is __________ .
(v) The probability of an event is greater than or equal to  and less than or equal to __________ .

Answer

(i) Probability of an event E + Probability of the event ‘not E’ = 1.
(ii) The probability of an event that cannot happen is 0. Such an event is called  impossible event.
(iii) The probability of an event that is certain to happen is 1. Such an event is called sure or certain event.
(iv) The sum of the probabilities of all the elementary events of an experiment is 1.
(v) The probability of an event is greater than or equal to 0 and less than or equal to 1.


2. Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.

Answer

(i) It does not have equally likely outcomes as it depends on various reasons like mechanical problems, fuels etc.
(ii) It does not have equally likely outcomes as it depends on the player how he/she shoots.
(iii) It has equally likely outcomes.
(iv)It has equally likely outcomes.

3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

Answer

Yes, tossing of a coin is a fair way of deciding which team should get the ball at the beginning of a football game because it has only two outcomes either head or tail. A coin is always unbiased.

4. Which of the following cannot be the probability of an event?
(A) 2/3     (B) -1.5     (C) 15%       (D) 0.7

Answer

The probability of an event is always greater than or equal to 0 and less than or equal to 1.
Thus, (B) -1.5 cannot be the probability of an event.

5. If P(E) = 0.05, what is the probability of ‘not E’?

Answer

P(E) = 0.05
also, P(E) + P(not E) = 1
⇒ P(not E) = 1 – P(E)
⇒ P(not E) = 1 – 0.05
⇒ P(not E) = 0.95

6. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?

Answer

(i) Since the bag contains only lemon flavoured.
Therefor, No. of orange flavoured candies = 0
Probability of taking out orange flavoured candies = 0/1 = 0

(ii) The bag only have lemon flavoured candies.
Probability of taking out lemon flavoured candies = 1/1 = 1

7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Answer

Let E be the event of having the same birthday.
P(E) = 0.992
⇒ P(E) + P(not E) = 1
⇒ P(not E) = 1 – P(E)
⇒ 1 – 0.992 = 0.008
The probability that the 2 students have the same birthday is 0.008

8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?

Answer

No. of red balls = 3
No. of black balls = 5
Total no. of balls = 5+3 = 8
(i) Probability of drawing red balls = No. of red balls/Total no. of balls = 3/8

(ii) Probability of drawing black balls = No. of black balls/Total no. of balls = 5/8

9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red ? (ii) white ? (iii) not green?

Answer

No. of red marbles = 5
No. of white marbles =8
No. of green marbles = 4
Total no. of balls = 5+8+4 = 17

(i) Favourable no. of elementary events = 5
Probability of taking out red marble = 5/17

(ii) Favourable no. of elementary events = 8
Probability of taking out red marble = 8/17

(iii) Favourable no. of elementary events = 4
Probability of taking out red marble = 4/17

Page No: 309

10. A piggy bank contains hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50 p coin ? (ii) will not be a ₹5 coin?

Answer

No. of 50p coins = 100
No. of ₹1 coins = 50
No. of ₹2 coins = 20
No. of ₹5 coins = 10
Total no. of coins = 100 + 50 + 20 + 10 = 180

(i) Favourable no. of elementary events = 100
Probability that it will be 50p coins = 100/180 = 5/9

(ii) Favourable no. of elementary events = 100+50+20 = 170
Probability that it will be 50p coins = 170/180 = 17/18

11. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (see Fig. 15.4). What is the probability that the fish taken out is a male fish?
Answer

No. of male fish in the tank = 5
no. of female fish in the tank = 8
Total number of fish in the tank = 5 + 8 = 13
Favourable number events = 5
Probability of taking out a male fish = 5/13

12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig. 15.5), and these are equally likely outcomes. What is the probability that it will point at
(i) 8 ?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?
Answer

Possible no. of events = 8
(i) Favourable number of events = 1
Probability that it will point at 8 = 1/8

(ii) Odd numbers = 1, 3, 5 and 7
Favourable number of events = 4
Probability that it will be an odd number = 4/8 = 1/2

(iii) Numbers greater than 2 = 3, 4, 5, 6, 7 and 8
Favourable number of events = 6
Probability that a number greater than 4 = 6/8 = 3/4

(iv) Numbers less than 9 = 1,2,3,4,5,6,7,8
Favourable number of events = 8
Probability that a number less than 9 = 8/8 = 1

13. A die is thrown once. Find the probability of getting
(i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number.

Answer

Possible numbers of events on throwing a dice = 6
Numbers on dice = 1,2,3,4,5 and 6

(i) Prime numbers = 2, 3 and 5
Favourable number of events = 3
Probability that it will be a prime number = 3/6 = 1/2

(ii) Numbers lying between 2 and 6 = 3, 4 and 5
Favourable number of events = 3
Probability that a number between 2 and 6 = 3/6 = 1/2

(iii) Odd numbers = 1, 3 and 5
Favourable number of events = 3
Probability that it will be an odd number = 3/6 = 1/2

14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the queen of diamonds

Answer

Possible numbers of events = 52

(i) Numbers of king of red colour = 2
Probability of getting a king of red colour = 2/52 = 1/26

(ii) Numbers of face cards = 12
Probability of getting a face card = 12/52 = 3/13

(iii) Numbers of red face cards = 6
Probability of getting a king of red colour = 6/52 = 3/26

(iv) Numbers of jack of hearts =1
Probability of getting a king of red colour = 1/52

(v) Numbers of king of spade = 13
Probability of getting a king of red colour = 13/52 = 1/4

(vi) Numbers of queen of diamonds = 1
Probability of getting a king of red colour = 1/52

15. Five cards the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

Answer

Total numbers of cards = 5

(i) Numbers of queen = 1
Probability of picking a queen = 1/5

(ii) When queen is drawn and put aside then total numbers of cards left is 4
(a) Numbers of ace = 1
Probability of picking an ace = 1/4
(a) Numbers of queen = 0
Probability of picking a queen = 0/4 = 0

16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Answer

Numbers of defective pens = 12
Numbers of good pens = 132
Total numbers of pen = 132 + 12 = 144 pens
Favourable number of events = 132
Probability of getting a good pen = 132/144 = 11/12

17. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Answer

(i) Total numbers of bulbs = 20
Numbers of defective bulbs = 4
Probability of getting a defective bulb = 4/20 = 1/5

(ii) One non defective bulb is drawn in (i) then the total numbers of bulb left is 19
Total numbers of events = 19
Favourable numbers of events =  19 – 4 = 15
Probability that the bulb is not defective = 15/19

18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.

Answer

Total numbers of discs = 50

(i) Total numbers of favourable events = 81
Probability that it bears a two-digit number = 81/90 = 9/10

(ii) Perfect square numbers = 1, 4, 9, 16, 25, 36, 49, 64 and 81
Favourable numbers of events = 9
Probability of getting a perfect square number = 9/90 = 1/10

(iii) Numbers which are divisible by 5 = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90
Favourable numbers of events = 18
Probability of getting a number divisible by 5 = 18/90 = 1/5

Page No: 310

19. A child has a die whose six faces show the letters as given below:
 
The die is thrown once. What is the probability of getting (i) A? (ii) D?
 
Answer
 
Total numbers of events = 6
 
(i) Total numbers of faces having A on it = 2
Probability of getting A = 2/6 = 1/3
 
(ii) Total numbers of faces having D on it = 1
Probability of getting A = 1/6
 
20. Suppose you drop a die at random on the rectangular region shown in Fig. 15.6. What is the probability that it will land inside the circle with diameter 1m?
Answer
 
Area of the rectangle = (3 × 2) m2 = 6m2
Area of the circle = πr2 = π(1/2)2 m2  = π/4 m2 
Probability that die will land inside the circle = (π/4) × 1/6 = π/24
 
21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) She will buy it?
(ii) She will not buy it?

Answer

Total numbers of pens = 144
Numbers of defective pens = 20
Numbers of non defective pens = 144 – 20 = 124

(i) Numbers of favourable events = 124
Probability that she will buy it = 124/144 = 31/36

(ii) Numbers of favourable events = 20
Probability that she will not buy it = 20/144 = 5/36

22. Refer to Example 13. (i) Complete the following table:
 
(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11. Do you agree with this argument? Justify your answer.
 
Answer

Events that can happen on throwing two dices are:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
Total numbers of events : 6 × 6 = 36

(i) To get sum as 2, possible outcomes = (1,1)
To get sum as 3, possible outcomes = (1,2) and (2,1)
To get sum as 4, possible outcomes = (1,3); (3,1);  and (2,2)
To get sum as 5, possible outcomes = (1,4); (4,1); (2,3);  and (3,2)
To get sum as 6, possible outcomes = (1,5); (5,1); (2,4); (4,2);  and (3,3)
To get sum as 7, possible outcomes = (1,6); (6,1); (5,2); (2,5); (4,3);  and (3,4)
To get sum as 8, possible outcomes = (2,6); (6,2); (3,5); (5,3);  and (4,4)
To get sum as 9, possible outcomes = (3,6); (6,3); (4,5);  and (5,4)
To get sum as 10, possible outcomes = (4,6); (6,4) and (5,5)
To get sum as 11, possible outcomes = (5,6) and (6,5)
To get sum as 12, possible outcomes = (6,6)

Event:
Sum on 2 dice		23456789101112
Probability  1/36   2/36   3/36   4/36   5/36   6/36   5/36   4/36   3/36   2/36   1/36 

(ii) No, i don’t agree with the argument. It is already justified in (i).

 23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Answer

Events that can happen in tossing 3 coins = HHH, HHT, HTH, THH, TTH, HTT, THT, TTT
Total number of events = 8
Hinif will lose the game if he gets HHT, HTH, THH, TTH, HTT, THT
Favourable number of elementary events = 6
Probability of losing the game = 6/8 = 3/4

24. A die is thrown twice. What is the probability that
(i) 5 will not come up either time? (ii) 5 will come up at least once?
[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

Answer

 (i) Consider the following events.
A = first throw shows 5,
B = second throw shows 5
P(A) = 6/36, P(B) = 6/36 and P(notB) = 5/6
⇒ P(notA) = 1– 6/36 = 30/36 = 5/6
Required probability = 5/6 × 5/6 = 25/36

(ii) Number of events when 5 comes at least once = 11
Probability = 11/36

Page No: 311

25. Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3
(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is 1/2

Answer

(i) The statement is incorrect
Possible events = (H,H); (H,T); (T,H) and (T,T)
Probability of getting two heads = 1/4
Probability of getting one of the each = 2/4 = 1/2

(ii) Correct. The two outcomes considered are equally likely.

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NCERT Solutions for Class 10 Maths Chapter 14 Statistics | EduGrown

In This Post we are  providing Chapter 14 Statistics NCERT Solutions for Class 10 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Statistics Class 10 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 10 Maths Statistics NCERT Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

Exercise 14.1

1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
Number of Plants   0-2       2-4       4-6      6-8       8-10       10-12       12-14   
Number of Houses    1215623
Which method did you use for finding the mean, and why?

Answer

No. of plants
   (Class interval)  		No. of houses (fi)Mid-point (xi)    fixi    
0-2111
2-4236
4-6155
6-85735
8-106954
10-1221122
12-1431339
 Sum fi = 20
    Sum fixi = 162    

Mean = x̄ = ∑fixi /∑fi = 162/20 = 8.1

We would use direct method because the numerical value of fi and xi are small.

2. Consider the following distribution of daily wages of 50 workers of a factory.
Daily wages (in Rs.)   100-120       120-140      140-160      160-180       180-200   
Number of workers    12148610
Find the mean daily wages of the workers of the factory by using an appropriate method.

Answer

Here, the value of mid-point (xi) is very large, so assumed mean A = 150 and class interval is h = 20.
So, ui = (xi – A)/h = ui = (xi – 150)/20

Daily wages
   (Class interval)  		Number of workers
frequency (fi)		Mid-point (xi)ui = (xi – 150)/20    fiui    
100-12012110-2-24
120-14014130-1-14
140-160815000
160-180617016
180-20010190220
TotalSum fi = 50
  Sum fiui = -12  
Mean = x̄ = A + h∑fiui /∑fi =150 + (20 × -12/50) = 150 – 4.8 = 145.20
Thus, mean daily wage = Rs. 145.20

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

Answer

Here, the value of mid-point (xi)  mean x̄ = 18

Class intervalNumber of children (fi)Mid-point (xi)    fixi    
11-1371284
13-1561484
15-17916144
17-191318 = A234
19-21f2020f
21-23522110
23-2542496
Totalfi = 44+f
 Sum fixi = 752+20f 

Mean = x̄ = ∑fixi /∑fi = (752+20f)/(44+f)
⇒ 18 = (752+20f)/(44+f)
⇒ 18(44+f) = (752+20f)
⇒ 792+18f = 752+20f
⇒ 792+18f = 752+20f
⇒ 792 – 752 = 20f – 18f
⇒ 40 = 2f
⇒ f = 20

Page No: 271

4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Answer

xi = (Upper limit + Lower limit)/2
Class size (h) = 3
Assumed mean (A) = 75.5

Class IntervalNumber of women (fi)Mid-point (xi)ui = (xi – 75.5)/hfiui
65-68266.5-3-6
68-71469.5-2-8
71-74372.5-1-3
74-77875.500
77-80778.517
80-83481.538
83-86284.536
 Sum fi= 30  Sum fiui = 4

Mean = x̄ = A + h∑fiui /∑fi = 75.5 + 3×(4/30) = 75.5 + 4/10 = 75.5 + 0.4 = 75.9
The mean heart beats per minute for these women is 75.9

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
 
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Answer

Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.45 from the lower limit.
Here, assumed mean (A) = 57
Class size (h) = 3

Class IntervalNumber of boxes (fi)Mid-point (xi)di = xi – Afidi
49.5-52.51551-690
52.5-55.511054-3-330
55.5-58.513557 = A00
58.5-61.5115603345
61.5-64.525636150
 Sum fi = 400  Sum fidi = 75

Mean = x̄ = A + ∑fidi /∑fi = 57 + (75/400) = 57 + 0.1875 = 57.19

6. The table below shows the daily expenditure on food of 25 households in a locality.
 
Find the mean daily expenditure on food by a suitable method.
 
Answer

Here, assumed mean (A) = 225

Class IntervalNumber of households (fi)Mid-point (xi)di = xi – Afidi
100-1504125-100-400
150-2005175-50-250
200-2501222500
250-300227550100
300-3502325100200
 Sum fi = 25  Sum fidi = -350

Mean = x̄ = A + ∑fidi /∑fi = 225 + (-350/25) = 225 – 14 = 211
The mean daily expenditure on food is 211

7. To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:
Find the mean concentration of SO2 in the air.

Answer
Concentration of SO2 (in ppm)Frequency (fi)Mid-point (xi)
fixi
0.00-0.0440.020.08
0.04-0.0890.060.54
0.08-0.1290.100.90
0.12-0.1620.140.28
0.16-0.2040.180.72
0.20-0.2420.200.40
TotalSum fi = 30 Sum (fixi) = 2.96

Mean = x̄ = ∑fixi /∑fi
= 2.96/30 = 0.099 ppm

Page No. 272

8. A class teacher has the following absentee record of 40 students of a class for the whole
term. Find the mean number of days a student was absent.
 

 
Number of days
0-66-1010-1414-2020-2828-3838-40
Number of students111074431

Answer

Class intervalFrequency (fi)
Mid-point (xi)
fixi
0-611333
6-1010880
10-1471284
14-2041768
20-2842496
28-3833399
38-4013939
 Sum fi = 40 Sum fixi = 499

Mean = x̄ = ∑fixi /∑fi
= 499/40 = 12.48 days

9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean
literacy rate.


Literacy rate (in %)45-5555-6565-7575-8585-98
Number of cities3101183

Answer

Class IntervalFrequency (fi)(xi)di = xi – aui = di/h
fiui
45-55350-20-2-6
55-651060-10-1-10
65-751170000
75-858801018
85-953902026
 Sum fi  = 35   Sum fiui  = -2

Mean = x̄ = a + (∑fiui /∑fi) х h
= 70 + (-2/35) х 10 = 69.42

Exercise 14.2

1. The following table shows the ages of the patients admitted in a hospital during a year:


Age (in years)5-1515-2525-3535-4545-5555-65
Number of patients6112123145
 


Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Answer

Modal class = 35 – 45, l = 35, class width (h) = 10, fm = 23, f1 = 21 and f2 = 14

 
Calculation of Mean:
 
Class IntervalFrequency (fi)Mid-point (xi)fixi
5-1561060
15-251120220
25-352130630
35-452340920
45-551450700
55-65560300
 Sum fi = 80 Sum fixi = 2830


Mean = x̄ = ∑fixi /∑fi
= 2830/80 = 35.37 yr

2. The following data gives the information on the observed lifetimes (in hours) of 225
electrical components :


Lifetime (in hours)0-2020-4040-6060-8080-100100-120
Frequency103552613829


Determine the modal lifetimes of the components.

Answer

Modal class of the given data is 60–80.
Modal class = 60-80, l = 60, fm = 61, f1 = 52, f2 = 38 and h = 20


3. The following data gives the distribution of total monthly household expenditure of 200
families of a village. Find the modal monthly expenditure of the families. Also, find the
mean monthly expenditure :


ExpenditureNumber of families
1000-150024
1500-200040
2000-250033
2500-300028
3000-350030
3500-400022
4000-450016
4500-50007


Answer

Modal class = 1500-2000, l = 1500, fm = 40, f1 = 24, f2 = 33 and h = 500

 
Calculation for mean:

 

Class Intervalfixidi = xi – aui = di/hfiui
1000-1500241250-1500-3-72
1500-2000401750-1000-2-80
2000-2500332250-500-1-33
2500-3000282750000
3000-3500303250500130
3500-40002237501000244
4000-45001642501500348
4500-5000747502000428
 fi = 200   fiui = -35


Mean = x̄ = a + (∑fiui /∑fi) х h
= 2750 + (35/200) х 500
= 2750 – 87.50 = 2662.50

 

 Question 4.
The following distribution gives the state-wise teacher- student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.2 Q4
Solution:
Chapter 14 Maths Class 10 NCERT Solutions Ex 14.2 PDF Download Q4
Chapter 14 Maths Class 10 NCERT Solutions Ex 14.2 PDF Download Q4.1

 

 Question 5.
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.2 Q5
Find the mode of the data.
Solution:
Statistics Class 10 Maths NCERT Solutions Ex 14.2 PDF Download Q5

 

Question 6.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.2 Q6
Solution:
Exercise 14.2 Class 10 Maths NCERT Solutions PDF Download Q6

 Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.3 Q1
Solution:
Statistics Class 10 Maths NCERT Solutions Ex 14.3 pdf download Q1
Statistics Class 10 Maths NCERT Solutions Ex 14.3 pdf download Q1.1

 

 Question 2.
If the median of the distribution given below is 28.5, find the values of x and y.
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.3 Q2
Solution:
Ex 14.3 Class 10 Maths NCERT Solutions pdf download Q2
Ex 14.3 Class 10 Maths NCERT Solutions pdf download Q2.1

 

 Question 3.
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.3 Q3
Solution:
Exercise 14.3 Class 10 Maths NCERT Solutions pdf download Q3

 

 Question 4.
The lengths of 40 leaves of a plant are measured correct to nearest millimetre, and the data obtained is represented in the following table:
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.3 Q4
Find the median length of the leaves.
Solution:
Exercise 14.3 Class 10 Maths NCERT Solutions pdf download Q4
Exercise 14.3 Class 10 Maths NCERT Solutions pdf download Q4.1

 

 Question 5.
The following table gives the distribution of the lifetime of 400 neon lamps:
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.3 Q5
Find the median lifetime of a lamp.
Solution:
Chapter 14 Maths Class 10 NCERT Solutions Ex 14.3 pdf download Q5

 

 Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabet in the surnames was obtained as follows:
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.3 Q6
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.
Solution:
Statistics Class 10 Maths NCERT Solutions Ex 14.3 pdf download Q6
Statistics Class 10 Maths NCERT Solutions Ex 14.3 pdf download Q6.1
Statistics Class 10 Maths NCERT Solutions Ex 14.3 pdf download Q6.2

 

Ex 14.3 Class 10 Maths Question 7.
The distribution below gives the weight of 30 students of a class. Find the median weight of the students.
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.3 Q7
Solution:
Statistics Class 10 Maths NCERT Solutions Ex 14.3 pdf download Q7

 Question 1.
The following distribution gives the daily income of 50 workers of a factory.
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.4 Q1
Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
Solution:
Statistics Class 10 Maths NCERT Solutions Ex 14.4 pdf download Q1
Statistics Class 10 Maths NCERT Solutions Ex 14.4 pdf download Q1.1

 

Ex 14.4 Class 10 Maths Question 2.
During the medical check-up of 35 students of a class, their weights were recorded as follows:
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.4 Q2
Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Solution:
Ex 14.4 Class 10 Maths NCERT Solutions pdf download
Ex 14.4 Class 10 Maths NCERT Solutions pdf download Q2.1

 

Question 3.
The following table gives production yield per hectare of wheat of 100 farms of a village.
NCERT Solutions For Class 10 Maths Chapter 14 Statistics Ex 14.4 Q3
Change the distribution to a more than type distribution, and draw its ogive.
Solution:
Exercise 14.4 Class 10 Maths NCERT Solutions pdf download Q3
Exercise 14.4 Class 10 Maths NCERT Solutions pdf download Q3.1

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NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes| EduGrown

In This Post we are  providing Chapter 13 Surface area & Volume NCERT Solutions for Class 10 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Surface area & Volume Class 10 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 10 Maths Surface area & Volume NCERT Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

NCERT Solutions for Class 10 Maths Chapter 13 Surface, Area & Volume

Exercise 13.1

Unless stated otherwise, take π = 22/7.

1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.

Answer

 Volume of each cube(a3) = 64 cm3
⇒ a3 = 64 cm3
⇒ a = 4 cm
Side of the cube = 4 cm
Length of the resulting cuboid = 4 cm
Breadth of the resulting cuboid = 4 cm
Height of the resulting cuboid = 8 cm
∴ Surface area of the cuboid = 2(lb + bh + lh)
= 2(8×4 + 4×4 + 4×8) cm2
= 2(32 + 16 + 32) cm2
= (2 × 80) cm2 = 160 cm2

2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

Answer

Diameter of the hemisphere = 14 cm
Radius of the hemisphere(r) = 7 cm
Height of the cylinder(h) = 13 – 7 = 6 cm
Also, radius of the hollow hemisphere = 7 cm
Inner surface area of the vessel = CSA of the cylindrical part + CSA of hemispherical part
= (2πrh+2πr2) cm2 = 2πr(h+r) cm2
= 2 × 22/7 × 7 (6+7) cm2 = 572 cm2

3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Answer

Radius of the cone and the hemisphere(r) = 3.5 = 7/2 cm
Total height of the toy = 15.5 cm
Height of the cone(h) = 15.5 – 3.5 = 12 cm

Curved Surface Area of cone = πrl = 22/7 × 7/2 × 25/2
= 275/2 cm2

Curved Surface Area of hemisphere = 2πr2
= 2 × 22/7 × (7/2)2
= 77 cm2

Total surface area of the toy = CSA of cone + CSA of hemisphere
= (275/2 + 77) cm2
= (275+154)/2 cm2 
= 429/2 cm2 = 214.5 cm2
The total surface area of the toy is 214.5 cm2

4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Answer

Each side of cube = 7 cm
∴ Radius of the hemisphere = 7/2 cm
Total surface area of solid = Surface area of cubical block + CSA of hemisphere – Area of base of hemisphere

TSA of solid = 6×(side)2 + 2πr2 – πr2
= 6×(side)2 + πr2
= 6×(7)2 + (22/7 × 7/2 × 7/2)
= (6×49) + (77/2)
= 294 + 38.5 = 332.5 cm2
The surface area of the solid is 332.5 cm2

5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Answer
Diameter of hemisphere = Edge of cube = l
Radius of hemisphere = l/2
Total surface area of solid = Surface area of cube + CSA of hemisphere – Area of base of hemisphere
TSA of remaining solid = 6 (edge)2 + 2πr2 – πr2
= 6l2 + πr2
= 6l2 + π(l/2)2
= 6l2 + πl2/4 
= l2/4 (24 + π) sq units

6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
 
Answer
 
Two hemisphere and one cylinder are given in the figure. 
Diameter of the capsule = 5 mm
∴ Radius = 5/2 = 2.5 mm
Length of the capsule = 14 mm
∴ Length of the cylinder = 14 – (2.5 + 2.5) = 9mm
Surface area of a hemisphere = 2πr2 = 2 × 22/7 × 2.5 × 2.5
= 275/7 mm2 

Surface area of the cylinder = 2πrh
= 2 × 22/7 × 2.5 x 9
= 22/7 × 45
990/7 mm2

∴ Required surface area of medicine capsule
= 2 × Surface area of hemisphere + Surface area of cylinder
= (2 × 275/7) × 990/7
= 550/7 + 990/7
= 1540/7 = 220 mm2

Page No. 245

7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m2. (Note that the base of the tent will not be covered with canvas.)

Answer

Tent is combination of cylinder and cone.
 
Diameter = 4 m
Slant height of the cone (l) = 2.8 m
Radius of the cone (r) = Radius of cylinder = 4/2 = 2 m
Height of the cylinder (h) = 2.1 m

∴ Required surface area of tent = Surface area of cone+Surface area of cylinder
= πrl + 2πrh
= πr(l+2h)
= 22/7 × 2 (2.8 + 2×2.1)
= 44/7 (2.8 + 4.2)
= 44/7 × 7 = 44 m2

Cost of the canvas of the tent at the rate of ₹500 per m2
= Surface area × cost per m2
= 44 × 500 = ₹22000

8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the
same height and same diameter is hollowed out. Find the total surface area of the
remaining solid to the nearest cm2.

Answer

Diameter of cylinder = diameter of conical cavity = 1.4 cm
∴ Radius of cylinder = Radius of conical cavity = 1.4/2 = 0.7
Height of cylinder = Height of conical cavity = 2.4 cm
TSA of remaining solid = Surface area of conical cavity+TSA of cylinder
= πrl + (2πrh + πr2)
= πr (l + 2h + r)
= 22/7 × 0.7 (2.5 + 4.8 + 0.7)
= 2.2 × 8 = 17.6 cm2
 
9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the total surface area of the article.
 
Answer
 
Given,
Height of the cylinder, h = 10cm and radius of base of cylinder = Radius of hemisphere (r) = 3.5 cm
Now, required total surface area of the article = 2 × Surface area of hemisphere + Lateral surface area of cylinder
Exercise 13.2
 
1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.
 
Answer
 
Given, solid is a combination of a cone and a hemisphere.
 
Also, we have radius of the cone (r) = Radius of the hemisphere = 1cm and height of the cone (h) = 1cm
∴ Required volume of the solid = Volume of the cone + Volume of the hemisphere
 
2. Rachel, an engineering, student was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel mode. (Assume the outer and inner dimensions of the model to be nearly the same.)
 
Answer
 
Given, model is a combination of a cylinder and two cones. Also, we have, diameter of the model,BC = ED = 3 cm.
 
3. A gulab jamun, contains sugar syrup upto about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see figure).
 
Answer
Let r be the radius of the hemisphere and cylinder both. h1 be the height of the hemisphere which is equal to its radius and h2 be the height of the cylinder. 
Given, length = 5 cm, diameter = 2.8 cm
 
4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see figure).
 
Answer
Given, length of cuboid (l) = 15 cm
Breadth of cuboid (b) = 10cm
and height of cuboid (h) = 3.5 cm
∴ Volume of cuboid = l×b×h = 15 ×10 × 3.5 = 5253
 
5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water upto the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel,one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
 
Answer
 
6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass. (Use π = 3.14)
Answer
 
7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
 
Answer
 
 
8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be
345 cm3. Check whether she is correct, taking the above as the inside measurements and π = 3.14.
 
Answer
 
She is not correct.
Exercise 13.3
 
1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
 
Answer
 
Given, the radius of the sphere (r) = 4.2 cm
Radius of the cylinder (r1) = 6 cm
 
2. Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
 
Answer
 
Let r1, r2 and r3 be the radius of given three spheres and R be the radius of a single solid sphere.
Given, r1 = 6 cm , r2 = 8 cm and r3 =10 cm
 
3. A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
 
Answer
 
Given, the height of deep well which form a cylinder (h1) = 20 m
 
4. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4m to form an embankment. Find the height of the embankment.
 
Answer
 
Given, the height of deep well which form a cylinder (h1) = 14 m
 
5. A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
 
Answer
 
Let the height and radius of ice cream container (cylinder) be h1 and r1.
 
6. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 10 3 5 . . cm cm cm  × ?
 
Answer
 
We know that, every coin has a shape of cylinder. Let radius and height of the coin are r1 and h1 respectively.
 
7. A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
 
Answer
 
Let the radius and slant height of the heap of sand are r and l.
Given, the height of the heap of sand h = 24 cm.
 
8. Water in a canal, 6 m wide and 1.5 m deep is flowing with a speed of 10 kmh−1. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
 
Answer
 
Given, speed of flow of water (l) = 10 kmh-1 = 10 × 1000 mh−1
Area of canal = 6 × 1.5 = 9m2
 
9. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 kmh−1, in how much time will the tank be filled?
 
Answer
 
Given, speed of flow of water = 3 kmh−1 = 3×1000 mh−1
∴ Length of water in 1 h = 3000 m
Exercise 13.4
 
1. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
 
Answer
 
Let the height of the frustum of a cone be h.
 
2. The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
 
Answer
 
Let the slant height of the frustum be l and radius of the both ends of the frustum be r1 and r2.
 
3. A fez, the cap used by the turks, is shaped like the frustum of a cone (see figure). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.
 
Answer
 
Let the slant height of fez be l and the radius of upper end which is closed be r1 and the other end which is open be r2.
 
4. A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of ` 20 per litre. Also find the cost of metal sheet used to make the container. If it costs ₹8 per 100 cm2. (Take π = 3.14)
 
Answer
 
Let h be the height of the container, which is in the form of a frustum of a cone whose lower end is closed and upper end is opened. Also, let the radius of its lower end be r1 and upper end be r2.
 
5. A metallic right circular cone 20 cm high and whose vertical angle is 60º is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.
 
Answer
 
Let r1 and r2 be the radii of the frustum of upper and lower ends cut by a plane. Given, height of the cone = 20 cm.
∴ Height of the frustum = 10 cm
 
 
Exercise 13.5 (Optional)
 
1. A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.
 
Answer
 
Since, the diameter of the wire is 3 mm.
When a wire is one round wound about a cylinder, it covers a 3 mm of length of the cylinder.
Given, length of the cylinder = 12 cm = 120 mm
∴ Number of rounds to cover 120 mm = 120/3 = 40
Given, diameter of a cylinder is d = 10 cm
∴ Radius, r = 10/2 = 5 cm
∴ Length of wire required to complete one round = 2πr = 2π(5) = 10π cm
∴ Length of the wire in covering the whole surface
= Length of the wire in completing 40 rounds
= 10π × 40 = 400π cm
= 400 × 3.14 = 1256 cm
Now, radius of copper wire = 3/2 mm = 3/20 cm
 
2. A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate).
 
Answer
 
Here, ABC is a right angled triangle at A and BC is the hypotenuse.
 
3. A cistern, internally measuring 150 cm × 120 cm× 110 cm, has 129600 cm3 of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. how many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?
 
Answer
 
Given, internally dimensions of cistern = 150 cm×120 cm×110 cm
∴ Volume of cistern = 150×120×110 
= 1980000 cm3
Volume of water = 129600 cm3
∴ Volume of cistern to be filled = 1980000 – 129600 = 1850400 cm3
Let required number of bricks = n
 
4. In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280 km2, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.
 
Answer
 
Given, area of the valley = 97280 km2
 
5. An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.
 
Answer
 
Given, oil funnel is a combination of a cylinder and a frustum of a cone.
Also, given height of cylindrical portion h = 10 cm
 
6. Derive the formula for the curved surface area and total surface area of the frustum of a cone. Using the symbols as explained.
 
Answer
 
Leth be the height, l be the slant height and r1 and r2 be the radii of the bases (r1>r2) of the frustum of a cone. We complete the conical part OCD.
The frustum of the right circular cone can be viewed as the difference of the two right circular cones OAB and OCD. Let slant height of the cone OAB be l1 and its height be h1 i.e.,OB = OA = l1 and OP = h1
The frustum of the right circular cone can be viewed as the difference of the two right circular cones OAB and OCD. Let slant height of the cone OAB be l1 and its height be h1 i.e., OB = OA = l1 and OP = h1
Then in ∆ACE,
 
7. Derive the formula for the volume of the frustum of a cone given to you in the section 13.5 using the symbols as explained.
 
Answer
 
Leth be the height, l be the slant height and r1 and r2 be the radii of the bases (r1>r2) of the frustum of a cone. We complete the conical part OCD.
The frustum of the right circular cone can be viewed as the difference of the two right circular cones OAB and OCD. Let the height of the cone OAB be h1 and its slant height be l1.
i.e., OP = h1 and OA = OB = l1
Then, height of the cone OCD = h1 −1
∆OQD ~ ∆OPB (AA similarity criterion)

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NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles | EduGrown

In This Post we are  providing Chapter 12 Area related to circle NCERT Solutions for Class 10 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Area related to circle Class 10 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 10 Maths Area related to circle NCERT Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

NCERT Solutions for Class 10 Maths Chapter 12 Area related to circle

Exercise: 12.1
 
Unless stated otherwise, use π =22/7.
 
1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
 
Answer

Let the radius of the third circle be R.
Circumference of the circle with radius R = 2πR
Circumference of the circle with radius 19 cm = 2π × 19 = 38π cm
Circumference of the circle with radius 9 cm = 2π × 9 = 18π cm
Sum of the circumference of two circles = 38π + 18π = 56π cm
Circumference of the third circle = 2πR = 56π
⇒ 2πR = 56π cm
⇒ R = 28 cm
The radius of the circle which has circumference equal to the sum of the circumferences of the two circles is 28 cm.
 
2.  The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.
 
Answer
 
Let the radius of the third circle be R.
Area of the circle with radius R = πR2
Area of the circle with radius 8 cm = π × 82 = 64π cm2
Area of the circle with radius 6 cm = π × 62 = 36π cm2
Sum of the area of two circles = 64π cm2 + 36π cm2 = 100π cm2
Area of the third circle = πR2 = 100π cm2
⇒ πR2 = 100π cm2
⇒ R2 = 100 cm2
⇒ R = 10 cm
Thus, the radius of the new circle is 10 cm.
 
3. Fig. 12.3 depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.
 
 
 
Answer

Diameter of Gold circle (first circle) = 21 cm
Radius of first circle, r1 = 21/2 cm = 10.5 cm
Each of the other bands is 10.5 cm wide,
∴ Radius of second circle, r2 = 10.5 cm + 10.5 cm = 21 cm
∴ Radius of third circle, r3 = 21 cm + 10.5 cm = 31.5 cm
∴ Radius of fourth circle, r4 = 31.5 cm + 10.5 cm = 42 cm
∴ Radius of fifth circle, r5 = 42 cm + 10.5 cm = 52.5 cm
Area of gold region = π r12 = π (10.5)2 = 346.5 cm2
Area of red region = Area of second circle – Area of first circle = π r22 – 346.5 cm2
                              = π(21)2 – 346.5 cm2 = 1386 – 346.5 cm2 = 1039.5 cm2

Area of blue region = Area of third circle – Area of second circle = π r32 – 1386 cm2

                                          =  π(31.5)2 – 1386 cm2 = 3118.5 – 1386 cm2 = 1732.5 cm2
 

Area of black region = Area of fourth circle – Area of third circle = π r32 – 3118.5 cm2

 

 

                                             = π(42)2 – 1386 cm2 = 5544 – 3118.5 cm2 = 2425.5 cm2

Area of white region = Area of fifth circle – Area of fourth circle = π r42 – 5544 cm2

                                             = π(52.5)2 – 5544 cm2 = 8662.5 – 5544 cm2 = 3118.5 cm2
 

4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?

 
Answer
 
Diameter of the wheels of a car = 80 cm
Circumference of wheels = 2πr = 2r × π = 80 π cm
Distance travelled by car in 10 minutes = (66 × 1000 × 100 × 10)/60 = 1100000 cm/s
No. of revolutions = Distance travelled by car/Circumference of wheels
                              = 1100000/80 π = (1100000 × 7)/(80×22) = 4375
4375 complete revolutions does each wheel make in 10 minutes.
 
5. Tick the correct answer in the following and justify your choice : If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
          (A) 2 units                     (B) π units                  (C) 4 units              (D) 7 units
 
Answer
 
Let the radius of the circle be r.
∴ Perimeter of the circle = Circumference of the circle = 2πr
∴ Area of the circle = π r2

A/q,

2πr = π r2
⇒ 2 = r
Thus, the radius of the circle is 2 units. (A) is correct.
Exercise: 12.2
 
Unless stated otherwise, use π =22/7.
1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
 
Answer
 
Area of the sector making angle θ = (θ/360°)×π r2 
Area of the sector making angle 60° = (60°/360°)×π r2 cm2
                         = (1/6)×62π  = 36/6 π cm2 = 6 × 22/7 cm2 = 132/7 cm2
 

2. Find the area of a quadrant of a circle whose circumference is 22 cm.

 
Answer
 
Quadrant of a circle means sector is making angle 90°.
Circumference of the circle = 2πr = 22 cm
Radius of the circle = r = 22/2π cm = 7/2 cm 
Area of the sector making angle 90° = (90°/360°)×π r2 cm2
                    = (1/4)×(7/2)2π  = (49/16) π cm2 = (49/16) × (22/7) cm2 = 77/8 cm2

3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
 
Answer
 
Here, Minute hand of clock acts as radius of the circle.
∴ Radius of the circle (r) = 14 cm
Angle rotated by minute hand in 1 hour = 360°
∴ Angle rotated by minute hand in 5 minutes = 360° × 5/60 = 30°
Area of the sector making angle 30° = (30°/360°)×π r2 cm2
                         = (1/12) × 142π  = 196/12 π cm2 = (49/3) × (22/7) cm2 = 154/3 cm2
Area swept by the minute hand in 5 minutes = 154/3 cm2

4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector. (Use π = 3.14)
 
Answer

Radius of the circle = 10 cm
Major segment is making 360° – 90° = 270°
Area of the sector making angle 270° 
                                 = (270°/360°) × π r2 cm2
                                           = (3/4) × 102π  = 75 π cm2
                                = 75 × 3.14 cm2 = 235.5 cm2
∴ Area of the major segment = 235.5 cm2
Height of ΔAOB = OA = 10 cm
Base of ΔAOB = OB = 10 cm
Area of ΔAOB = 1/2 × OA × OB
                         = 1/2 ×10 × 10 = 50 cm2
Major segment is making  90°
 Area of the sector making angle 90°                                                                
= (90°/360°) × π r2 cm2
= (1/4) × 102π  = 25 π cm2
 = 25 × 3.14 cm2 = 78.5 cm2
Area of the minor segment = Area of the sector making angle 90° – Area of ΔAOB
                                            = 78.5 cm2 –  50 cm2 = 28.5 cm2
 

5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
   (i) the length of the arc
   (ii) area of the sector formed by the arc
   (iii) area of the segment formed by the corresponding chord

 
Answer
 
Radius of the circle = 21 cm
 
(i) Length of the arc AB = θ/360° × 2πr
                                        = 60°/360° × 2 × 22/7 × 21
                                        = 1/6 × 2 × 22/7 × 21 = 22
The length of the arc is  22 cm.
 
(ii) Angle subtend by the arc = 60° 
Area of the sector making angle 60° = (60°/360°) × π r2 cm2
                                          = (1/6) × 212π  = 441/6 π cm2
                               = 441/6 × 22/7 cm2 = 231 cm2
∴ Area of the sector formed by the arc is 231 cm2

(iii) Area of equilateral ΔAOB = √3/4 × (OA)2 = √3/4 × 212 = (441√3)/4 cm2
Area of the segment formed by the corresponding chord 
                     = Area of the sector formed by the arc – Area of equilateral ΔAOB
                     = 231 cm2 – (441√3)/4 cm2 
 

6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and √3 = 1.73)

 
Answer
 
 
Radius of the circle = 15 cm
ΔAOB is isosceles as two sides are equal.
∴ ∠A = ∠B
Sum of all angles of triangle = 180°
∠A + ∠B + ∠C = 180°
⇒ 2 ∠A = 180° – 60°
⇒ ∠A = 120°/2
⇒ ∠A = 60°
Triangle is equilateral as ∠A = ∠B = ∠C = 60°
∴ OA = OB = AB = 15 cm
Area of equilateral ΔAOB = √3/4 × (OA)2 = √3/4 × 152 
                                          = (225√3)/4 cm2 = 97.3 cm2

Angle subtend at the centre by minor segment = 60°
Area of Minor sector making angle 60° = (60°/360°) × π r2 cm2
                                                                                     = (1/6) × 152 π  cm2 =  225/6 π  cm2
                                                  =  (225/6) × 3.14 cm2 = 117.75  cm2
 

Area of the minor segment = Area of Minor sector – Area of equilateral ΔAOB

                                            = 117.75  cm2 – 97.3 cm2 = 20.4 cm2


Angle made by Major sector = 360° – 60° = 300°
Area of the sector making angle 300° = (300°/360°) × π r2 cm2
                                                   = (5/6) × 152 π  cm2 =  1125/6 π  cm2
                                                  =  (1125/6) × 3.14 cm2 = 588.75  cm2
 

Area of major segment = Area of Minor sector + Area of equilateral ΔAOB

                                            = 588.75  cm2 + 97.3 cm2 = 686.05 cm2
 

7. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73)

 
Answer

 
Radius of the circle, r = 12 cm
Draw a perpendicular OD to chord AB. It will bisect AB.
∠A = 180° – (90° + 60°) = 30°
cos 30° = AD/OA
⇒ √3/2 = AD/12
⇒ AD = 6√3 cm
⇒ AB = 2 × AD = 12√3 cm
sin 30° = OD/OA
⇒ 1/2 = OD/12
⇒ OD = 6 cm
Area of ΔAOB = 1/2 × base × height
                         = 1/2 × 12√3 × 6 = 36√3 cm
                         = 36 × 1.73 = 62.28 cm2

Angle made by Minor sector = 120°
Area of the sector making angle 120° = (120°/360°) × π r2 cm2
                                                   = (1/3) × 122 π  cm2 =  144/3 π  cm2
                                                  =  48 × 3.14 cm2 = 150.72  cm2
∴ Area of the corresponding Minor segment = Area of the Minor sector – Area of ΔAOB
                                                                        = 150.72  cm2 – 62.28 cm2
                                                                                                = 88.44 cm2

8. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m
long rope (see Fig. 12.11). Find
 (i) the area of that part of the field in which the horse can graze.
(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)
 
Answer
 
Side of square field = 15 m
Length of rope is the radius of the circle, r = 5 m
Since, the horse is tied at one end of square field, it will graze only quarter of the field with radius 5 m.
 
(i) Area of circle = π r2 = 3.14 × 52 = 78.5 m2
Area of that part of the field in which the horse can graze = 1/4 of area of the circle = 78.5/4 = 19.625 m2
 

(ii) Area of circle if the length of rope is increased to 10 m = π r2 = 3.14 × 102 = 314 m2

Area of that part of the field in which the horse can graze = 1/4 of area of the circle 
                                                                                            = 314/4 = 78.5 m2

Increase in grazing area = 78.5 m2 – 19.625 m2 = 58.875 m2

 

 

9. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. 12.12. Find:

(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.
 
Answer
 
Number of diameters = 5
Length of diameter = 35 mm
∴ Radius = 35/2 mm

(i) Total length of silver wire required = Circumference of the circle +  Length of 5 diameter
                                                              = 2π r + (5×35) mm = (2 × 22/7 × 35/2) + 175 mm
                                                              = 110 + 175 mm = 185 mm

(ii) Number of sectors = 10
Area of each sector = Total area/Number of sectors
Total Area = π r2 = 22/7 × (35/2)2 = 1925/2 mm2
∴ Area of each sector = (1925/2) × 1/10 = 385/4 mm2

10. An umbrella has 8 ribs which are equally spaced (see Fig. 12.13). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
 
 
Answer
 
Number of ribs in umbrella = 8
Radius of umbrella while flat = 45 cm
Area between the two consecutive ribs of the umbrella =
                                            Total area/Number of ribs
Total Area = π r2 = 22/7 × (45)2 = 6364.29 cm2
∴ Area between the two consecutive ribs = 6364.29/8 cm2
                                                                   = 795.5 cm2


11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.
 
Answer
 
Angle of the sector of circle made by wiper = 115°
Radius of wiper = 25 cm
Area of the sector made by wiper = (115°/360°) × π r2 cm2
                                                                        = 23/72 × 22/7 × 252 =  23/72 × 22/7 × 625 cm2
                                                                        = 158125/252 cm2
 
Total area cleaned at each sweep of the blades = 2 ×158125/252 cm2 =  158125/126 = 1254.96 cm2
 

12. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned.

(Use π = 3.14)
 
Answer
 
Let O bet the position of Lighthouse.
Distance over which light spread i.e. radius, r = 16.5 km
Angle made by the sector = 80°
Area of the sea over which the ships are warned = Area made by the sector.
Area of sector = (80°/360°) × π r2 km2
                               = 2/9 × 3.14 × (16.5)2 km2
                               =  189.97 km2


 
 
 
13. A round table cover has six equal designs as shown in Fig. 12.14. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm2 . (Use √3 = 1.7)

Answer


Number of equal designs = 6
Radius of round table cover = 28 cm
Cost of making design = ₹ 0.35 per cm2
∠O = 360°/6 = 60°

ΔAOB is isosceles as two sides are equal. (Radius of the circle)
∴ ∠A = ∠B
Sum of all angles of triangle = 180°
∠A + ∠B + ∠O = 180°
⇒ 2 ∠A = 180° – 60°
⇒ ∠A = 120°/2
⇒ ∠A = 60°
Triangle is equilateral as ∠A = ∠B = ∠C = 60°
Area of equilateral ΔAOB = √3/4 × (OA)2 = √3/4 × 282 = 333.2 cm2

Area of sector ACB = (60°/360°) × π r2 cm2
                                            = 1/6 × 22/7 × 28 × 28 = 410.66 cm2
Area of design = Area of sector ACB – Area of equilateral ΔAOB
                        = 410.66 cm2 – 333.2 cm2 = 77.46 cm2
 

Area of 6 design = 6 × 77.46 cm2 = 464.76 cm2



Total cost of making design = 464.76 cm2 × ₹ 0.35 per cm2 = ₹ 162.66
 
14. Tick the correct answer in the following :
Area of a sector of angle p (in degrees) of a circle with radius R is
  (A) p/180 × 2πR             (B) p/180 × π R2                   (C) p/360 × 2πR           (D) p/720 × 2πR2 
 
Answer
 
Area of a sector of angle p = p/360 × π R2
                                                         = p/360 × 2/2 × π R2 
                                           =  2p/720 × 2πR2 

Hence, Option (D) is correct.

Exercise: 12.3
 
Unless stated otherwise, use π =22/7
1. Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.
 
Answer 
 
PQ = 24 cm and PR = 7 cm
∠P = 90° (Angle in the semi-circle)
∴ QR is hypotenuse of the circle = Diameter of the circle.
By Pythagoras theorem,
QR2 = PR2  + PQ2 
⇒ QR2 = 72  + 242
⇒ QR2 = 49  + 576
⇒ QR2 = 625
⇒ QR = 25 cm
∴ Radius of the circle = 25/2 cm
Area of the semicircle = (π R2)/2
                                    = (22/7 × 25/2 × 25/2)/2 cm2
                                                = 13750/56 cm2 = 245.54 cm2
Area of the ΔPQR = 1/2 × PR × PQ
                              = 1/2 × 7 × 24 cm2
                                        = 84 cm2
Area of the shaded region = 245.54 cm2 – 84 cm2 = 161.54 cm2
 
2. Find the area of the shaded region in Fig. 12.20, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.
 
Answer
 
Radius inner circle = 7 cm
Radius of outer circle = 14 cm
Angle made by sector = 40°
Area of the sector OAC = (40°/360°) × π r2 cm2
                                                = 1/9 × 22/7 × 142 = 68.44 cm2
Area of the sector OBD = (40°/360°) × π r2 cm2
                                                                       = 1/9 × 22/7 × 72  = 17.11 cm2
 
Area of the shaded region ABDC = Area of the sector OAC – Area of the sector circle OBD
                                                      = 68.44 cm2 – 17.11 cm2 = 51.33 cm2
 
3. Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side 14 cm and APD and BPC are semicircles.


Answer
 
There are two semicircles in the figure.
Side of the square = 14 cm
Diameter of the semicircle = 14 cm
∴ Radius of the semicircle = 7 cm
Area of the square = 14 × 14 = 196 cm2
Area of the semicircle = (π R2)/2
                                    = (22/7 × 7 × 7)/2 cm2 = 77 cm2 
Area of two semicircles = 2 × 77 cm2  = 154 cm2
 
Area of the shaded region = 196 cm2 – 154 cm2 = 42 cm2
 
4. Find the area of the shaded region in Fig. 12.22, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.
 
Answer
 
OAB is an equilateral triangle with each angle equal to 60°.
Area of the sector is common in both.
Radius of the circle = 6 cm.
Side of the triangle = 12 cm.
Area of the equilateral triangle = √3/4 × (OA)2 = √3/4 × 122 = 36√3 cm2
Area of the circle = π R2 = 22/7 × 62 = 792/7 cm2
Area of the sector making angle 60° = (60°/360°) × π r2 cm2
                                                                              = 1/6 × 22/7 × 62 cm2  = 132/7 cm2
Area of the shaded region = Area of the equilateral triangle + Area of the circle – Area of the sector
                                          = 36√3 cm2 + 792/7 cm2 – 132/7 cm2
                                                        = (36√3 + 660/7) cm2 
 
5. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in Fig. 12.23. Find the area of the remaining portion of the square.
 
Answers 
 
Side of the square = 4 cm
Radius of the circle = 1 cm
Four quadrant of a circle are cut from corner and one circle of radius are cut from middle.
Area of square = (side)2 = 42 = 16 cm2
Area of the quadrant = (π R2)/4 cm2 = (22/7 × 12)/4 = 11/14 cm2
∴ Total area of the 4 quadrants = 4 × (11/14) cm2 = 22/7 cm2
Area of the circle = π R2 cm2 = (22/7 × 12) = 22/7 cm2
Area of the shaded region = Area of square – (Area of the 4 quadrants + Area of the circle)
                                           = 16 cm2 – (22/7 + 22/7) cm2
                                                         = 68/7 cm2
 
6. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. 12.24. Find the area of the design.


Answer
 
Radius of the circle = 32 cm
Draw a median AD of the triangle passing through the centre of the circle.
⇒ BD = AB/2
Since, AD is the median of the triangle
∴ AO = Radius of the circle = 2/3 AD
⇒ 2/3 AD = 32 cm
⇒ AD = 48 cm
In ΔADB,
By Pythagoras theorem,
AB2 = AD2  + BD2 
⇒ AB2 = 482  + (AB/2)2
⇒ AB2 = 2304  + AB2/4
⇒ 3/4 (AB2) = 2304
⇒ AB2 = 3072
⇒ AB = 32√3 cm
Area of ΔADB = √3/4 × (32√3)2 cm2 = 768√3 cm2
Area of circle = π R2 = 22/7 × 32 × 32 = 22528/7 cm2

Area of the design = Area of circle – Area of ΔADB
                              = (22528/7 – 768√3) cm2
 
Page No: 236
 
7. In Fig. 12.25, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

Answer

Side of square = 14 cm
Four quadrants are included in the four sides of the square.
∴ Radius of the circles = 14/2 cm = 7 cm
Area of the square ABCD = 142 = 196 cm2
Area of the quadrant = (π R2)/4 cm2 = (22/7 × 72)/4 cm2
                                             = 77/2 cm2
Total area of the quadrant = 4 × 77/2 cm2 = 154 cm2 
 
Area of the shaded region = Area of the square ABCD – Area of the quadrant
                                          = 196 cm2 – 154 cm2
                                                        = 42 cm2 
 
8. Fig. 12.26 depicts a racing track whose left and right ends are semicircular.
The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If
the track is 10 m wide, find :
(i) the distance around the track along its inner edge
(ii) the area of the track.
 
Answer
 
Width of track = 10 m
Distance between two parallel lines = 60 m
Length of parallel tracks = 106 m
DE = CF = 60 m
Radius of inner semicircle, r = OD = O’C 
                                               = 60/2 m = 30 m
Radius of outer semicircle, R = OA = O’B
                                               = 30 + 10 m = 40 m
Also, AB = CD = EF = GH = 106 m
 
 
Distance around the track along its inner edge = CD + EF + 2 × (Circumference of inner semicircle)
                                                                          = 106 + 106 + (2 × πr) m = 212 + (2 × 22/7 × 30) m
                                                                          = 212 + 1320/7 m = 2804/7 m
 
Area of the track = Area of ABCD + Area EFGH + 2 × (area of  outer semicircle) – 2 × (area of  inner  semicircle)
                           = (AB × CD) + (EF × GH) + 2 × (πr2/2) – 2 × (πR2/2) m2
                           = (106 × 10) + (106 × 10) + 2 × π/2 (r2 -R2) m2
                           = 2120 + 22/7 × 70 × 10 m2
                                 = 4320 m2 
 
9. In Fig. 12.27, AB and CD are two diameters of a circle (with centre O) perpendicular to each other
and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Answer

Radius of larger circle, R = 7 cm
Radius of smaller circle, r = 7/2 cm
Height of ΔBCA = OC = 7 cm
Base of ΔBCA = AB = 14 cm
Area of ΔBCA = 1/2 × AB × OC = 1/2 × 7 × 14 = 49 cm2 
Area of larger circle = πR2 = 22/7 × 72 = 154 cm2 
Area of larger semicircle = 154/2 cm2 = 77 cm2 
Area of smaller circle = πr2 = 22/7 × 7/2 × 7/2 = 77/2 cm2
 
Area of the shaded region = Area of larger circle – Area of triangle – Area of larger semicircle + Area   of smaller circle
Area of the shaded region = (154 – 49 – 77 + 77/2) cm2
                                                        = 133/2 cm2 = 66.5 cm2 
 
10. The area of an equilateral triangle ABC is 17320.5 cm2. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see Fig. 12.28). Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205)
 
Answer
 
ABC is an equilateral triangle.
∴ ∠A = ∠B = ∠C = 60°
There are three sectors each making 60°.
Area of ΔABC = 17320.5 cm2
⇒ √3/4 × (side)2  = 17320.5
⇒ (side)2  = 17320.5 × 4/1.73205
⇒ (side)2  = 4 × 104 
⇒ side  = 200 cm
Radius of the circles = 200/2 cm = 100 cm
Area of the sector = (60°/360°) × π r2 cm2
                                       = 1/6 × 3.14 × (100)2 cm2
                                       = 15700/3 cm2
Area of 3 sectors = 3 × 15700/3 = 15700 cm2 =

Area of the shaded region = Area of equilateral triangle ABC – Area of 3 sectors
                                          = 17320.5 – 15700 cm2 = 1620.5 cm2
 
11. On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig. 12.29). Find the area of the remaining portion of the handkerchief.
 
Answer
Number of circular design = 9
Radius of the circular design = 7 cm
There are three circles in one side of square handkerchief.
∴ Side of the square = 3 × diameter of circle = 3 × 14 = 42 cm
Area of the square = 42 × 42 cm2 = 1764 cm2
 
Area of the circle = π r2 = 22/7 × 7 × 7 = 154 cm2
Total area of the design = 9 × 154 = 1386 cm2
 
Area of the remaining portion of the handkerchief = Area of the square – Total area of the design
                                                                                 = 1764 – 1386 = 378 cm2
 
12. In Fig. 12.30, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the
 (i) quadrant OACB,                       (ii) shaded region.
Answer
 
Radius of the quadrant = 3.5 cm = 7/2 cm
 
(i) Area of quadrant OACB = (πR2)/4 cm2
                                             = (22/7 × 7/2 × 7/2)/4 cm2
                                                            = 77/8 cm2
(ii) Area of triangle BOD = 1/2 × 7/2 × 2 cm2
                                                       = 7/2 cm2
Area of shaded region = Area of quadrant – Area of triangle BOD
                                    = (77/8 – 7/2) cm2 = 49/8 cm2 
                                                = 6.125 cm2
 
13. In Fig. 12.31, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)
 
Answer
Side of square = OA = AB = 20 cm
Radius of the quadrant = OB
OAB is right angled triangle
By Pythagoras theorem in ΔOAB ,
OB2 = AB2  + OA2 
⇒ OB2 = 202  + 202
⇒ OB2 = 400  + 400
⇒ OB2 = 800
⇒ OB = 20√2 cm
Area of the quadrant = (πR2)/4 cm2 = 3.14/4 × (20√2)2 cm2 = 628 cm2
Area of the square = 20 × 20 = 400  cm2
 
Area of the shaded region = Area of the quadrant – Area of the square
                                          = 628 – 400 cm2 = 228 cm2
 
14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Fig. 12.32). If ∠AOB = 30°, find the area of the shaded region.
 
Answer
 
Radius of the larger circle, R = 21 cm
Radius of the smaller circle, r = 7 cm
Angle made by sectors of both concentric circles =  30°
Area of the larger sector = (30°/360°) × π R2 cm2
                                                     = 1/12 × 22/7 × 212 cm2
                                                     = 231/2 cm2
Area of the smaller circle = (30°/360°) × π r2 cm2
                                                     = 1/12 × 22/7 × 72 cm2
                                                     = 77/6 cm2
 
Area of the shaded region = 231/2 – 77/6 cm2
                                                        = 616/6 cm2 = 308/3 cm2
 
15. In Fig. 12.33, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC
as diameter. Find the area of the shaded region.
 
Answer
 
Radius of the the quadrant ABC of circle = 14 cm
AB = AC = 14 cm
BC is diameter of semicircle.
ABC is right angled triangle.
By Pythagoras theorem in ΔABC,
BC2 = AB2  + AC2 
⇒ BC2 = 142  + 142
⇒ BC = 14√2 cm
Radius of semicircle = 14√2/2 cm = 7√2 cm

Area of ΔABC = 1/2 × 14 × 14 = 98 cm2
Area of quadrant = 1/4 × 22/7 × 14 × 14 = 154 cm2
Area of the semicircle = 1/2 × 22/7 × 7√2 × 7√2 = 154 cm2
 
Area of the shaded region = Area of the semicircle + Area of ΔABC – Area of quadrant
                                          = 154 + 98 – 154 cm2 = 98 cm2

Page No: 238
 
16. Calculate the area of the designed region in Fig. 12.34 common between the two quadrants of circles of radius 8 cm each.
 
Answer
AB = BC = CD = AD = 8 cm
Area of ΔABC = Area of ΔADC = 1/2 × 8 × 8 = 32 cm2
Area of quadrant AECB = Area of quadrant AFCD = 1/4 × 22/7 × 82
                                                     = 352/7 cm2
 
Area of shaded region = (Area of quadrant AECB – Area of ΔABC) +                                           (Area of quadrant AFCD – Area of ΔADC)
                                    = (352/7 – 32) + (352/7 -32) cm2
                                                = 2 × (352/7 -32) cm2
                                                 =  256/7 cm2

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NCERT Solutions for Class 10 Maths Chapter 11 Construction | EduGrown

In This Post we are  providing Chapter 11 Construction NCERT Solutions for Class 10 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Construction Class 10 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 10 Maths Construction NCERT Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

NCERT Solutions for Class 10 Maths Chapter 11 Construction

Exercise 11.1

In each of the following, give the justification of the construction also:

1. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.

Answer

Steps of Construction:
Step I: AB = 7.6 cm is drawn.
Step II: A ray AX making an acute angle with  AB is drawn.
Step III: After that, a ray BY is drawn parallel to AX making an equal acute angle as in the previous step.
Step IV: Point A1, A2, A3, A4 and A5 is marked on AX and point B1, B2…. to B8 is marked on BY such that AA1 = A1A2 = A2A3 =….BB1= B1B2 = …. B7B8
Step V: A5 and B8 are joined and it intersected AB at point C diving it in the ratio 5:8.


AC : CB = 5 : 8

Justification:
ΔAA5C ~ ΔBB8C
∴ AA5/BB8 = AC/BC = 5/8

 
2.  Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.
 
Answer

Steps of Construction:
Step I: AB = 6 cm is drawn.
Step II: With A as a centre and radius equal to 4cm, an arc is draw.
Step III: Again, with B as a centre and radius equal to 5 cm an arc is drawn on same  side of AB intersecting previous arc at C.
Step IV: AC and BC are joined to form ΔABC.
Step V: A ray AX is drawn making an acute  angle with AB below it.
Step VI: 5 equal points (sum of the ratio = 2 + 3 =5) is marked on AX as A1 A2….A5
Step VII: A5B is joined. A2B’ is drawn parallel to A5B and B’C’ is drawn parallel to BC.
ΔAB’C’ is the required triangle
 

Justification:
∠A(Common)
∠C = ∠C’ and ∠B = ∠ B’ (corresponding angles)
Thus ΔAB’C’ ~  ΔABC by AAA similarity condition
From the figure,
AB’/AB = AA2/AA5 = 2/3
AB’ =2/3 AB
AC’ = 2/3 AC

3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are
7/5 of the corresponding sides of the first triangle.

Answer

Steps of Construction:
Step I: A triangle ABC with sides 5 cm, 6 cm and 7 cm is drawn.
Step II: A ray BX making an acute angle with BC is drawn opposite to vertex A.
Step III: 7 points as B1 B2 B3 B4 B5 B6 and B7 are marked on BX.
Step IV; Point B5 is joined with C to draw B5C.
Step V: B7C’ is drawn parallel to B5C and C’A’ is parallel to CA.
Thus A’BC’ is the required triangle.

Justification
ΔAB’C’ ~  ΔABC by AAA similarity condition
∴ AB/A’B = AC/A’C’ = BC/BC’
and BC/BC’ = BB5/BB7 = 5/7
∴A’B/AB = A’C’/AC = = BC’/BC = 7/5

4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1.5 times the corresponding sides of the isosceles triangle.

Answer

Steps of Construction:
Step I: BC = 5 cm is drawn.
Step II: Perpendicular bisector of BC is drawn and it intersects BC at O.
Step III: At a distance of 4 cm, a point A is marked on the perpendicular bisector of BC.
Step IV: AB and AC is joined to form ΔABC.
Step V: A ray BX is drawn making an acute angle with BC opposite to vertex A.
Step VI: 3 points B1 B2 and B3 is marked BX.
Step VII: B2 is joined with C to form B2C.
Step VIII: B3C’ is drawn parallel to B2C and C’A’ is drawn parallel to CA.
Thus, A’BC’ is the required triangle formed.

Justification:

ΔAB’C’ ~  ΔABC by AA similarity condition.
∴ AB/AB’ = BC/B’C’ = AC/AC’
also,
AB/AB’ = AA2/AA3 = 2/3
⇒ AB’ = 3/2 AB, B’C’ = 3/2 BC and AC’ = 3/2 AC
 
 
5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.
 
Answer

Steps of Construction:
Step I: BC = 6 cm is drawn.
Step II: At point B, AB = 5 cm is drawn making an
            ∠ABC = 60° with BC.
Step III: AC is joined to form ΔABC.
Step IV: A ray BX is drawn making an acute angle with BC  opposite to vertex A.
Step V: 4 points B1 B2 B3 and B4 at equal distance is marked on BX.
Step VII: B3 is joined with C’ to form B3C’.
Step VIII: C’A’ is drawn parallel CA.
Thus, A’BC’ is the required triangle.

Justification:

∠A = 60° (Common)
∠C = ∠C’
ΔAB’C’ ~  ΔABC by AA similarity condition.
∴ AB/AB’ = BC/B’C’ = AC/AC’
also,
AB/AB’ = AA3/AA4 = 4/3
⇒ AB’ = 3/4 AB, B’C’ = 3/4 BC and AC’ = 3/4 AC

6. Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of Δ ABC.

Answer

Sum of all side of triangle = 180°
∴ ∠A  + ∠B  + ∠C  = 180°
∠C = 180° – 150° = 30°
Steps of Construction:
Step I: BC = 7 cm is drawn.
Step II: At B, a ray is drawn making an angle of  45° with BC.
 
Step III: At C, a ray making an angle of 30° with  BC is drawn intersecting the previous ray at A. Thus, ∠A = 105°.
Step IV: A ray BX is drawn making an acute angle with BC opposite to vertex A.
Step V: 4 points B1 B2 B3 and B4 at equal distance is marked on BX.
Step VI: B3C is joined and B4C’ is made parallel to B3C.
Step VII: C’A’ is made parallel CA.
Thus, A’BC’ is the required triangle.

Justification:

∠B = 45° (Common)
∠C = ∠C’
ΔAB’C’ ~  ΔABC by AA similarity condition.
∴ BC/BC’ = AB/A’B’ = AC/A’C’
also,
BC/BC’ = BB3/BB4 = 34
⇒ AB = 4/3 AB’, BC = 4/3 BC’ and AC = 4/3 A’C’

7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.

Answer

Steps of Construction:
Step I: BC = 3 cm is drawn.
Step II: At B, A ray making an angle of 90° with BC is drawn.
Step III: With B as centre and radius equal to 4 cm, an arc is made on the previous ray intersecting it at point A.
Step IV: AC is joined to form ΔABC.
Step V: A ray BX is drawn making an acute angle with BC  opposite to vertex A.
Step VI: 5 points B1 B2 B3 B4 and B5 at equal distance is marked on BX.
Step VII: B3C is joined B5C’ is made parallel to B3C.
Step VIII: A’C’ is joined together.
Thus, ΔA’BC’ is the required triangle.

Justification:
As in the previous question 6
Exercise 11.2

In each of the following, give also the justification of the construction:

1.  Draw a circle of a radius of 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.

Answer

Steps of Construction:
Step I: With O as a centre and radius equal to 6 cm, a circle is drawn.
Step II: A point P at a distance of 10 cm from the centre O is taken. OP is joined.
Step III: Perpendicular bisector OP is drawn and let it intersected at M.
Step IV: With M as a centre and OM as a radius, a circle is drawn intersecting the previous circle at Q and R.
Step V: PQ and PR are joined.
Thus, PQ and PR are the tangents to the circle.
On measuring the length, tangents are equal to 8 cm.
PQ = PR = 8cm.

Justification:
OQ is joined.
∠PQO = 90° (Angle in the semi-circle)
∴ OQ ⊥ PQ
Therefore, OQ is the radius of the circle then PQ has to be a tangent of the circle. 
Similarly, PR is a tangent of the circle.
 
2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
 
Answer
 
Steps of Construction:
Step I: With O as a centre and radius equal to 4 cm, a circle is drawn.
Step II: With O as a centre and radius equal to 6 cm, a concentric circle is drawn.
Step III: P be any point on the circle of radius 6 cm and OP is joined.
Step IV: Perpendicular bisector of OP is drawn which cuts it at                 M
 
Step V: With M as a centre and OM as a radius, a circle is  drawn which intersect the circle of radius 4 cm at Q and R
Step VI: PQ and PR are joined.
Thus, PQ and PR are the two tangents.

Measurement:
OQ = 4 cm (Radius of the circle)
PQ = 6 cm ( Radius of the circle)
∠PQO = 90° (Angle in the semi circle)
Applying Pythagoras theorem in ΔPQO,
PQ2 + QO2 = PO2
⇒ PQ2 + 42= 62
⇒ PQ2 + 16 = 36
⇒ PQ2 = 36 – 16
⇒ PQ2 = 20
⇒ PQ = 2√5

Justification:

∠PQO = 90° (Angle in the semi circle)
∴ OQ ⊥ PQ
Therefor, OQ is the radius of the circle then PQ has to be a tangent of the circle. 
Similarly, PR is a tangent of the circle.

3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.

Answer

Steps of Construction:
Step I: With O as a centre and radius equal to 3 cm, a                   circle is drawn.
Step II: The diameter of the circle is extended both sides and an arc is made to cut it at 7 cm.
Step III: Perpendicular bisector of OP and OQ is drawn and x and y be its mid-point.
Step IV: With O as a centre and Ox be its radius, a circle is drawn which intersected the previous circle at M and N.
Step V: Step IV is repeated with O as centre and Oy as radius and it intersected the circle at R and T.
Step VI: PM and PN are joined also QR and QT are joined.
Thus,  PM and PN are tangents to the circle from P and QR and QT are tangents to the circle from point Q.

Justification:

∠PMO = 90° (Angle in the semi-circle)
∴ OM ⊥ PM
Therefore, OM is the radius of the circle then PM has to be a tangent of the circle. 
Similarly, PN, QR and QT are tangents of the circle.

4. Draw a pair of tangents to a circle of radius 5 cm which is inclined to each other at an angle of 60°.

Answer

We know that radius of the circle is perpendicular to the tangents.
Sum of all the 4 angles of quadrilateral = 360°
∴ Angle between the radius (∠O)  = 360° – (90° + 90° + 60°) = 120°
Steps of Construction:
Step I: A point Q is taken on the circumference of the circle and OQ is joined. OQ is the radius of the circle.
Step II: Draw another radius OR making an angle equal to 120° with the previous one.
Step III: A point P is taken outside the circle. QP and PR are joined which is perpendicular OQ and OR.
Thus, QP and PR are the required tangents inclined to each other at an angle of 60°.

Justification:

Sum of all angles in the quadrilateral PQOR = 360°
∠QOR + ∠ORP + ∠OQR + ∠RPQ = 360°
⇒ 120° + 90° + 90° + ∠RPQ = 360°
⇒∠RPQ = 360° – 300°
⇒∠RPQ = 60°
Hence, QP and PR are tangents inclined to each other at an angle of 60°.

5. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.

Answer

Steps of Construction:
Step I: A line segment AB of 8 cm is drawn.
Step II: With A as centre and radius equal to 4 cm, a circle is drawn which cut the line at point O.
Step III: With B as a centre and radius equal to  3 cm, a circle is drawn.
Step IV: With O as a centre and OA as a   radius, a circle is drawn which intersect the previous two circles at P, Q and R, S.
Step V: AP, AQ, BR and BS are joined.
Thus, AP, AQ, BR and BS are the required tangents.

Justification:
∠BPA = 90° (Angle in the semi circle)
∴ AP ⊥ PB
Therefor, BP is the radius of the circle then AP has to be a tangent of the circle. 
Similarly,  AQ, BR and BS are tangents of the circle.

6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.

Answer

Steps of Construction:
Step I: A ΔABC is drawn.
Step II: Perpendicular to AC is drawn to point B which intersected it at D.
Step III: With O as a centre and OC as a radius, a   circle is drawn. The circle through B, C,D is drawn.
Step IV: OA is joined and a circle is drawn with diameter OA which intersected the previous circle at B and E.
Step V: AE is joined.
Thus, AB and AE are the required tangents to the circle from A.

Justification:

∠OEA = 90° (Angle in the semi-circle)
∴ OE ⊥ AE
Therefore, OE is the radius of the circle then AE has to be a tangent of the circle. 
Similarly,  AB is the tangent of the circle.

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NCERT Solutions for Class 10 Maths Chapter 10 Circle | EduGrown

In This Post we are  providing Chapter 8 Circle NCERT Solutions for Class 10 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Circle Class 10 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 10 Maths Circle NCERT Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

NCERT Solutions for Class 10 Maths Chapter 10 Circle

Exercise: 10.1
 
1. How many tangents can a circle have?
 
Answer
 
A circle can have infinite tangents.
 
2.  Fill in the blanks :
(i) A tangent to a circle intersects it in …………… point(s).
(ii) A line intersecting a circle in two points is called a ………….
(iii) A circle can have …………… parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called …………
 
Answer
 
(i) one
(ii) secant
(iii) two
(iv) point of contact
 
3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at
a point Q so that OQ = 12 cm. Length PQ is :
(A) 12 cm
(B) 13 cm
(C) 8.5 cm 
(D) √119 cm
 
Answer
 

The line drawn from the centre of the circle to the tangent is perpendicular to the tangent.
∴ OP ⊥ PQ

By Pythagoras theorem in ΔOPQ,
OQ2 = OP2 + PQ2
⇒ (12)2 = 52 + PQ2
⇒PQ2 = 144 – 25
⇒PQ2 = 119
⇒PQ = √119 cm
(D) is the correct option.
 
4. Draw a circle and two lines parallel to a given line such that one is a tangent and the
other, a secant to the circle.
 
Answer
 
AB and XY are two parallel lines where AB is the tangent to the circle at point C while XY is the secant to the circle.
Exercise: 10.2
 
In Q.1 to 3, choose the correct option and give justification.
 
1.  From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(A)  7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
 
Answer

The line drawn from the centre of the circle to the tangent is perpendicular to the tangent.
∴ OP ⊥ PQ
also, ΔOPQ is right angled.
OQ = 25 cm and PQ = 24 cm (Given)
By Pythagoras theorem in ΔOPQ,
OQ2 = OP2 + PQ2
⇒ (25)2 = OP2 + (24)2
⇒ OP2 = 625 – 576
⇒ OP2 = 49
⇒ OP = 7 cm
The radius of the circle is option (A) 7 cm.


2.  In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to
(A) 60°
(B) 70° 
(C) 80° 
(D) 90°
 
Answer

OP and OQ are radii of the circle to the tangents TP and TQ respectively.
∴ OP ⊥ TP and,
∴ OQ ⊥ TQ
∠OPT = ∠OQT = 90°
In quadrilateral POQT,
Sum of all interior angles = 360°
∠PTQ + ∠OPT + ∠POQ + ∠OQT  = 360°
⇒ ∠PTQ + 90° + 110° + 90°  = 360°
⇒ ∠PTQ = 70°
∠PTQ is equal to option (B) 70°.
 
3.  If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠ POA is equal to
(A) 50°  
(B) 60° 
(C) 70°
(D) 80°

Answer
 
OA and OB are radii of the circle to the tangents PA and PB respectively.
∴ OA ⊥ PA and,
∴ OB ⊥ PB
∠OBP = ∠OAP = 90°
In quadrilateral AOBP,
Sum of all interior angles = 360°
∠AOB + ∠OBP + ∠OAP + ∠APB  = 360°
⇒ ∠AOB + 90° + 90° + 80°  = 360°
⇒ ∠AOB = 100°
Now,
In ΔOPB and ΔOPA,
AP = BP (Tangents from a point are equal)
OA = OB (Radii of the circle)
OP = OP (Common side)
∴ ΔOPB ≅ ΔOPA (by SSS congruence condition)
Thus ∠POB = ∠POA
∠AOB = ∠POB + ∠POA
⇒ 2 ∠POA = ∠AOB
⇒ ∠POA = 100°/2 = 50°
∠POA is equal to option  (A) 50°
 
Page No: 214
 
4.  Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
 
Answer
 
Let AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A and B respectively.
Radii of the circle to the tangents will be perpendicular to it.
∴ OB ⊥ RS and,
∴ OA ⊥ PQ
∠OBR = ∠OBS = ∠OAP = ∠OAQ = 90º
From the figure,
∠OBR = ∠OAQ (Alternate interior angles)
∠OBS = ∠OAP (Alternate interior angles)
Since alternate interior angles are equal, lines PQ and RS will be parallel.
Hence Proved that the tangents drawn at the ends of a diameter of a circle are parallel.
 
5.  Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
 
Answer
 
Let AB be the tangent to the circle at point P with centre O.
We have to prove that PQ passes through the point O.
Suppose that PQ doesn’t passes through point O. Join OP.
Through O, draw a straight line CD parallel to the tangent AB.
PQ intersect CD at R and also intersect AB at P.
AS, CD // AB PQ is the line of intersection,
∠ORP = ∠RPA (Alternate interior angles)
but also,
∠RPA = 90° (PQ ⊥ AB) 
⇒ ∠ORP  = 90°
∠ROP + ∠OPA = 180° (Co-interior angles)
⇒∠ROP + 90° = 180°
⇒∠ROP = 90°
Thus, the ΔORP has 2 right angles i.e. ∠ORP  and ∠ROP which is not possible.
Hence, our supposition is wrong. 
∴ PQ passes through the point O.
 
6.  The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
 
Answer
 
AB is a tangent drawn on this circle from point A.
∴ OB ⊥ AB
OA = 5cm and AB = 4 cm (Given)
In ΔABO,
By Pythagoras theorem in ΔABO,
OA2 = AB2 + BO2
⇒ 52 = 42 + BO2
⇒ BO2 = 25 – 16
⇒ BO2 = 9
⇒ BO = 3
∴ The radius of the circle is 3 cm.

7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the
larger circle which touches the smaller circle.

Answer

Let the two concentric circles with centre O.
AB be the chord of the larger circle which touches the smaller circle at point P. 
∴ AB is tangent to the smaller circle to the point P.
 
⇒ OP ⊥ AB
By Pythagoras theorem in ΔOPA,
OA2 =  AP2 + OP2
⇒ 52 = AP2 + 32
⇒ AP2 = 25 – 9
⇒ AP = 4
In ΔOPB,
Since OP ⊥ AB,
AP = PB (Perpendicular from the center of the circle bisects the chord)
AB = 2AP = 2 × 4 = 8 cm
 ∴ The length of the chord of the larger circle is 8 cm.

8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC

Answer


From the figure we observe that,
DR = DS (Tangents on the circle from point D) … (i)
AP = AS (Tangents on the circle from point A) … (ii)
BP = BQ (Tangents on the circle from point B) … (iii)
CR = CQ (Tangents on the circle from point C) … (iv)
Adding all these equations,
DR + AP + BP + CR = DS + AS + BQ + CQ
⇒ (BP + AP) + (DR + CR)  = (DS + AS) + (CQ + BQ)
⇒ CD + AB = AD + BC
 
9. In Fig. 10.13, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠ AOB = 90°.
 
Answer
 
We joined O and C
A/q,
In ΔOPA and ΔOCA,
OP = OC (Radii of the same circle)
AP = AC (Tangents from point A)
AO = AO (Common side)
∴ ΔOPA ≅ ΔOCA (SSS congruence criterion)
⇒ ∠POA = ∠COA … (i)
Similarly,
 ΔOQB  ≅ ΔOCB
∠QOB = ∠COB … (ii)
Since POQ is a diameter of the circle, it is a straight line.
∴ ∠POA + ∠COA + ∠COB + ∠QOB = 180 º
From equations (i) and (ii),
2∠COA + 2∠COB = 180º
⇒ ∠COA + ∠COB = 90º
⇒ ∠AOB = 90°
 
10.  Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
 
Answer
Consider a circle with centre O. Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends ∠AOB at center O of the circle.
It can be observed that
OA ⊥ PA
∴ ∠OAP = 90°
Similarly, OB ⊥ PB
∴ ∠OBP = 90°
In quadrilateral OAPB,
Sum of all interior angles = 360º
∠OAP +∠APB +∠PBO +∠BOA = 360º
⇒ 90º + ∠APB + 90º + ∠BOA = 360º
⇒ ∠APB + ∠BOA = 180º
∴ The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
 
11. Prove that the parallelogram circumscribing a circle is a rhombus.
 
Answer
 
ABCD is a parallelogram,
∴ AB = CD … (i)
∴ BC = AD … (ii)
From the figure, we observe that,
DR = DS (Tangents to the circle at D)
CR = CQ (Tangents to the circle at C)
BP = BQ (Tangents to the circle at B)
AP = AS (Tangents to the circle at A)
Adding all these,
DR + CR + BP + AP = DS + CQ + BQ + AS
⇒ (DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
⇒ CD + AB = AD + BC … (iii)
Putting the value of (i) and (ii) in equation (iii) we get,
2AB = 2BC
⇒ AB = BC … (iv)
By Comparing equations (i), (ii), and (iv) we get,
AB = BC = CD = DA
∴ ABCD is a rhombus.
 
12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.
 
Answer
 
In ΔABC,
Length of two tangents drawn from the same point to the circle are equal,
∴ CF = CD = 6cm
∴ BE = BD = 8cm
∴ AE = AF = x
We observed that,
AB = AE + EB = x + 8
BC = BD + DC = 8 + 6 = 14
CA = CF + FA = 6 + x
Now semi perimeter of triangle (s) is,
⇒ 2s = AB + BC + CA
= x + 8 + 14 + 6 + x
= 28 + 2x
⇒s = 14 + x
Area of ΔABC = √s (s – a)(s – b)(s – c)
 
= √(14 + x) (14 + x – 14)(14 + x – x – 6)(14 + x – x – 8)
= √(14 + x) (x)(8)(6)
= √(14 + x) 48 x … (i)
also, Area of ΔABC = 2×area of (ΔAOF + ΔCOD + ΔDOB)
= 2×[(1/2×OF×AF) + (1/2×CD×OD) + (1/2×DB×OD)]
= 2×1/2 (4x + 24 + 32) = 56 + 4x … (ii)
Equating equation (i) and (ii) we get,
√(14 + x) 48 x = 56 + 4x
Squaring both sides,
48x (14 + x) = (56 + 4x)2
⇒ 48x = [4(14 + x)]2/(14 + x)
⇒ 48x = 16 (14 + x)
⇒ 48x = 224 + 16x
⇒ 32x = 224
⇒ x = 7 cm
Hence, AB = x + 8 = 7 + 8 = 15 cm
CA = 6 + x = 6 + 7 = 13 cm
 
13.  Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. 
 
Answer

Let ABCD be a quadrilateral circumscribing a circle with O such that it touches the circle at point P, Q, R, S. Join the vertices of the quadrilateral ABCD to the center of the circle.
In ΔOAP and ΔOAS,
AP = AS (Tangents from the same point)
OP = OS (Radii of the circle)
OA = OA (Common side)
ΔOAP ≅ ΔOAS (SSS congruence condition)
∴ ∠POA = ∠AOS
⇒∠1 = ∠8
Similarly we get,
∠2 = ∠3
∠4 = ∠5
∠6 = ∠7
Adding all these angles,
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 +∠8 = 360º
⇒ (∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360º
⇒ 2 ∠1 + 2 ∠2 + 2 ∠5 + 2 ∠6 = 360º
⇒ 2(∠1 + ∠2) + 2(∠5 + ∠6) = 360º
⇒ (∠1 + ∠2) + (∠5 + ∠6) = 180º
⇒ ∠AOB + ∠COD = 180º
Similarly, we can prove that ∠ BOC + ∠ DOA = 180º
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

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NCERT Solutions for Class 10 Maths Chapter 9 Some Application of Trigonometry | EduGrown

In This Post we are  providing Chapter 8 Some Application Of Trigonometry NCERT Solutions for Class 10 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These  Introduction to Trigonometry  Class 10 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 10 Maths Some Application Of Trigonometry NCERT Written Solutions  & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

NCERT Solutions for Class 10 Maths Chapter 9 Some Application Of Trigonometry

Exercise: 9.1
 
Page No: 203
 
 
1. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see Fig. 9.11). 
 
Answer

Let AB be the vertical pole Ac be 20 m  long rope tied to point C.
In  right ΔABC,
sin 30° = AB/AC
⇒ 1/2 = AB/20
⇒ AB = 20/2
⇒ AB = 10
The height of the pole is 10 m.
 
2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
 
Answer
 
Let AC be the broken part of the tree.
∴ Total height of the tree = AB+AC
In  right ΔABC,
cos 30° = BC/AC
⇒ √3/2 = 8/AC
⇒ AC = 16/√3
Also,
tan 30° = AB/BC
⇒ 1/√3 = AB/8
⇒ AB = 8/√3
Total height of the tree = AB+AC = 16/√3 + 8/√3 = 24/√3
 
3. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
 
Answer
 
There are two slides of height 1.5 m and 3 m. (Given)
Let AB is 1.5 m and PQ be 3 m slides. 
ABC is the slide inclined at 30° with length AC and PQR is the slide inclined at 
60° with length PR.
A/q,
In  right ΔABC,
sin 30° = AB/AC
⇒ 1/2 = 1.5/AC
⇒ AC = 3m
also,
In  right ΔPQR,
sin 60° = PQ/PR
⇒ √3/2 = 3/PR
⇒ PR = 2√3 m
Hence, length of the slides are 3 m and 2√3 m respectively.
 
Page No: 204
 
4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
 
Answer
 
Let AB be the height of the tower and C is the point elevation which is 30 m away from the foot of the tower.
A/q,
In  right ΔABC,
tan 30° = AB/BC
⇒ 1/√3 = AB/30
⇒ AB = 10√3
Thus, the height of the tower is 10√3 m.

5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Answer

Let BC be the height of the kite from the ground,
AC be the inclined length of the string from the ground and A is the point where string of the kite is tied.
A/q,
In  right ΔABC,
sin 60° = BC/AC
⇒ √3/2 = 60/AC
⇒ AC = 40√3 m
Thus, the length of the string from the ground is 40√3 m.

6.  A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Answer

Let the boy initially standing at point Y with inclination 30° and then he approaches the building to
the point X with inclination 60°.
∴ XY is the distance he walked towards the building.
also, XY = CD.
Height of the building = AZ = 30 m
AB = AZ – BZ = (30 – 1.5) = 28.5 m
A/q,
In  right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = 28.5/BD
⇒ BD = 28.5√3 m
also,
In  right ΔABC,
tan 60° = AB/BC
⇒ √3 = 28.5/BC
⇒ BC = 28.5/√3 = 28.5√3/3 m
∴ XY = CD = BD – BC = (28.5√3 – 28.5√3/3) = 28.5√3(1-1/3) = 28.5√3 × 2/3 = 57/√3 m.
Thus, the distance boy walked towards the building is 57/√3 m.

7.  From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Answer

Let BC be the 20 m high building.
D is the point on the ground from where the elevation is taken.
Height of transmission tower = AB = AC – BC
A/q,
In  right ΔBCD,
tan 45° = BC/CD
⇒ 1 = 20/CD
⇒ CD = 20 m
also,
In  right ΔACD,
tan 60° = AC/CD
⇒ √3 = AC/20
⇒ AC = 20√3 m
Height of transmission tower = AB = AC – BC = (20√3 – 20) m = 20(√3 – 1) m.

8.  A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Answer

Let AB be the height of statue.
D is the point on the ground from where the elevation is taken.
Height of pedestal = BC = AC – AB
A/q,
In  right ΔBCD,
tan 45° = BC/CD
⇒ 1 =  BC/CD
⇒ BC = CD.
also,
In  right ΔACD,
tan 60° = AC/CD
⇒ √3 = AB+BC/CD
⇒ √3CD = 1.6 m + BC
⇒ √3BC = 1.6 m + BC
⇒ √3BC – BC = 1.6 m
⇒ BC(√3-1) = 1.6 m
⇒ BC = 1.6/(√3-1) m
⇒ BC = 0.8(√3+1) m
Thus, the height of the pedestal is 0.8(√3+1) m.

9. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Answer

Let CD be the height of the tower equal to 50 m (Given)
Let AB be the height of the building.
BC be the distance between the foots of the building and the tower.
Elevation is 30° and 60° from the tower and the building respectively.
A/q,
In  right ΔBCD,
tan 60° = CD/BC
⇒ √3 = 50/BC
⇒ BC = 50/√3
also,
In  right ΔABC,
tan 30° = AB/BC
⇒ 1/√3 = AB/BC
⇒ AB = 50/3
Thus, the height of the building is 50/3.

10. Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

Answer

Let AB and CD be the poles of equal height.
O is the point between them from where the height of elevation taken.
BD is the distance between the poles.
A/q,
AB = CD,
OB + OD = 80 m
Now,
In  right ΔCDO,
tan 30° = CD/OD
⇒ 1/√3 = CD/OD
⇒ CD = OD/√3 … (i)
also,
In  right ΔABO,
tan 60° = AB/OB
⇒ √3 = AB/(80-OD)
⇒ AB = √3(80-OD)
AB = CD (Given)
⇒ √3(80-OD) = OD/√3
⇒ 3(80-OD) = OD
⇒ 240 – 3 OD = OD
⇒ 4 OD = 240
⇒ OD = 60
Putting the value of OD in equation (i)
CD = OD/√3 ⇒ CD = 60/√3 ⇒ CD = 20√3 m
also,
OB + OD = 80 m ⇒ OB = (80-60) m = 20 m
Thus, the height of the poles are 20√3 m and distance from the point of elevation are 20 m and 60 m respectively.


11.  A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.

Answer

Here, AB is the height of the tower.
CD = 20 m (given)
A/q,
In  right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = AB/(20+BC)
⇒ AB = (20+BC)/√3 … (i)
also,
In  right ΔABC,
tan 60° = AB/BC
⇒ √3 = AB/BC
⇒ AB = √3 BC … (ii)
From eqn (i) and (ii)
AB = √3 BC = (20+BC)/√3
⇒ 3 BC = 20 + BC
⇒ 2 BC = 20 ⇒ BC = 10 m
Putting the value of BC in eqn (ii)
AB = 10√3 m
Thus, the height of the tower 10√3 m and the width of the canal is 10 m.

12.  From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Answer

Let AB be the building of height 7 m and EC be the height of tower.
A is the point from where elevation of tower is 60° and the angle of depression of its foot is 45°
EC = DE + CD
also, CD = AB = 7 m.
and BC = AD
A/q,
In  right ΔABC,
tan 45° = AB/BC
⇒ 1= 7/BC
⇒ BC = 7 m = AD
also,
In  right ΔADE,
tan 60° = DE/AD
⇒ √3 = DE/7
⇒ DE = 7√3 m
Height of the tower = EC =  DE + CD
                                = (7√3 + 7) m = 7(√3+1) m.

13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Answer


Let AB be the lighthouse of height 75 m.
Let C and D be the positions of the ships.
30° and 45° are the angles of depression from the lighthouse.
A/q,
In  right ΔABC,
tan 45° = AB/BC
⇒ 1= 75/BC
⇒ BC = 75 m
also,
In  right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = 75/BD
⇒ BD = 75√3  m
 
The distance between the two ships = CD = BD – BC = (75√3 – 75) m = 75(√3-1) m.

Page No: 205

 
14.  A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m
from the ground. The height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see Fig. 9.13). Find the distance travelled by the balloon during the interval.

Answer

Let the initial position of the balloon be A and final position be B.
Height of balloon above the girl height = 88.2 m – 1.2 m = 87 m
Distance travelled by the balloon =
 DE = CE – CD
A/q,
In  right ΔBEC,
tan 30° = BE/CE
⇒ 1/√3= 87/CE
⇒ CE = 87√3 m
also,
In  right ΔADC,
tan 60° = AD/CD
⇒ √3= 87/CD
⇒ CD = 87/√3 m = 29√3 m
Distance travelled by the balloon =  DE = CE – CD = (87√3 – 29√3) m = 29√3(3 – 1) m = 58√3 m.

15.  A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Answer

Let AB be the tower.
D is the initial and C is the final position of the car respectively.
Angles of depression are measured from A.
BC is the distance from the foot of the tower to the car.
A/q,
In  right ΔABC,
tan 60° = AB/BC
⇒ √3 = AB/BC
⇒ BC = AB/√3 m
also,
In  right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = AB/(BC + CD)
⇒ AB√3 = BC + CD
⇒ AB√3 = AB/√3 + CD
⇒ CD = AB√3 – AB/√3
⇒ CD = AB(√3 – 1/√3)
⇒ CD = 2AB/√3
Here, distance of BC is half of CD. Thus, the time taken is also half.
Time taken by car to travel distance CD = 6 sec.
Time taken by car to travel BC = 6/2 = 3 sec.

16.  The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Answer

Let AB be the tower.
C and D be the two points with distance 4 m and 9 m from the base respectively.
A/q,
In  right ΔABC,
tan x = AB/BC
⇒ tan x = AB/4
⇒ AB = 4 tan x … (i)
also,
In  right ΔABD,
tan (90°-x) = AB/BD
⇒ cot x = AB/9
⇒ AB = 9 cot x  … (ii)
Multiplying  eqn (i) and (ii)
AB2 = 9 cot x × 4 tan x
⇒ AB2 = 36
⇒ AB = ± 6
Height cannot be negative. Therefore, the height of the tower is 6 m. Hence, Proved.

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